Bare-bones outline of eigenvalue theory and the Jordan canonical form

Size: px
Start display at page:

Download "Bare-bones outline of eigenvalue theory and the Jordan canonical form"

Transcription

1 Bare-bones outline of eigenvalue theory and the Jordan canonical form April 3, 2007 N.B.: You should also consult the text/class notes for worked examples. Let F be a field, let V be a finite-dimensional vector space over F, and let T : V V be a linear operator. 1. Definition of eigenvalues and eigenvectors. An element λ F is called an eigenvalue of T if there exists a nonzero vector x V such that T x = λx; such a vector is called an eigenvector (or λ-eigenvector) of T. Clearly, we have the equivalences λ is an eigenvalue of T ker(t λi) 0 T λi is not injective det(t λi) = 0, because we know that a linear transformation has nonzero kernel if and only if it is not invertible if and only if it has zero determinant. In particular, 0 is an eigenvalue of T if and only if T is not invertible. The λ-eigenspace of T is defined to be the set of λ-eigenvectors of T : it is a subspace of V. 2. Eigenvalues exist when F is algebraically closed. First, recall that F is algebraically closed if every polynomial p(x) = a 0 + a 1 X + + a n X n, with a i F, has a zero in F, i.e., there exists c F such that p(c) = 0. If the ground field F is algebraically closed, then we can prove the existence of an eigenvalue of T, as follows. First proof. Since F is algebraically closed, the polynomial χ(x) = det(t XI) has a root, which is necessarily an eigenvalue of T (see 1). Second proof (longer, but more elementary). Let x V be any nonzero vector. Because dim V <, there is a smallest d 1 such that the vectors x, T x,..., T d x are linearly dependent. Thus, there are scalars a 0,..., a d, not all zero, such that 0 = a 0 x + a 1 T x + + a d T d x = p(t )x, (1) where p(x) is the polynomial p(x) = a 0 +a 1 X + +a d X d. Since F is algebraically closed, we can factor p(x) into linear factors, say p(x) = a d (X c d ) (X c 1 ) with c i F. The minimality property defining d implies that a d 0. So from Eq. (1) we get 0 = (T c d )(T c d 1 ) (T c 1 )x. From this equality it follows that one of the c i must be an eigenvalue of T : If (T c 1 )x = 0 then c 1 is an eigenvalue of T, otherwise there must be some i > 1 for which v = (T c i 1 ) (T c 1 )x 0, but (T c i )v = 0. Page 1 of 9

2 Thus, in principle, the problem of finding eigenvalues is straightforward: write down the polynomial det(t XI) and find its roots. In practice, however, the best one can usually do is determine the roots approximately. There is also the matter of computing the polynomial det(t XI), which is a computationally expensive task. 3. Dimension bounds the number of eigenvalues. If dim V = n, then there are at most n distinct eigenvalues of T. More precisely: If λ 1,..., λ k are distinct eigenvalues of T, then any corresponding set of eigenvectors x 1,..., x k is linearly independent. Proof. We, of course, start by writing a 1 x a k x k = 0 for scalars a i F, and try to show that all the a i s are necessarily zero. We will show that a 1 = 0, by finding an operator S such that Sx 1 0 but Sx i = 0 for i > 1. Applying S to the zero-sum a 1 x a k x k will then give 0 = a 1 Sx 1, thereby forcing a 1 = 0. The vanishing of the other coefficients follows from the same techinque (for different choices of the operator S). Note this simple technique carefully, for we will use it several times in these notes, and it is fairly standard in linear algebra. Let S be the linear operator (T λ 2 ) (T λ n ). The factors of S commute so the equality (T λ 2 )x 2 = 0 implies that Sx 2 = 0, and the same line of reasoning shows that Sx i = 0 for all i 1. On the other hand, upon applying S to x 1, we get Sx 1 = i 1 (λ 1 λ i )x 1, which is not zero because the eigenvalues λ i are pairwise unequal. We deduce that a 1 = 0, as explained in the previous paragraph. To complete the proof, it remains to apply the same technique to x 2, x 3, etc., in the obvious way. 4. Counterexamples. Eigenvalues might not exist if F fails to be algebraically closed. For example, the field F = R is not algebraically closed, and any non-trivial rotation about the origin of R 2 is a real-linear transformation with no (real) eigenvalues, as no nozero vector of R 2 is scaled under rotation. The definition of an eigenvalue makes sense when V is infinite-dimensional, but then their existence is not guaranteed, even if F is algebraically closed. For example, if V were the infinite-dimensional vector space C[X] of complex polynomials, then the operator T = d/dx: V V has no eigenvector: the exponential function is an infinite power series, and is therefore not an element of V. 5. Diagonalizability. The operator T is said to be diagonalizable if there is a basis V consisting of eigenvectors of T ; with respect to such a basis, the matrix of T is a diagonal matrix whose diagonal elements are the eigenvalues of T. Not every operator (or, equivalently, matrix) is diagonalizable. For example, the rational matrix T = ( ) is not diagonalizable: the only eigenvalue of T is 1, and clearly the 1-eigenvectors of T cannot span Q 2 (as T I). The main question that we will be concerned with in these notes is: When is T diagonalizable, and how can one tell? If T is diagonalizable, then certainly T possesses sufficiently many eigenvalues. In Section 2 we saw that the existence of eigenvalues is guaranteed when F is algebraically closed (and don t forget our standing assumption that dim V < ), so we shall often assume this. Under this assumption, our question will be resolved by the existence of the so-called Jordan canonical form of T ( 20). Page 2 of 9

3 6. Generalized eigenvectors, multiplicity of eigenvalues. Let λ F. A vector x V is called a generalized (λ-)eigenvector if (T λ) k x = 0 for some k 1. The set of generalized λ-eigenvectors form a vector subspace of V. We write V T (λ) = {The space of generalized λ-eigenvectors of T }. Clearly, this space is nonzero if and only if λ is an eigenvalue of T. The multiplicity of λ is defined to be the dimension of V T (λ). [Note: In the literature, generalized eigenvectors are sometimes called root vectors. ] The impossibility of diagonalizing a general operator T is an expression of the fact that V is not, in general, spanned by the eigenspaces of T (e.g., there is only one eigenspace of T = ( ) as an operator of Q 2, and it is 1-dimensional). This deficiency of eigenvectors is corrected by the notion of a generalized eigenvector. 7. Main theorem 1: Generalized eigenspace decomposition. If F is algebraically closed, then V = λ F V T (λ). More specifically, if λ 1,..., λ k is the set of (distinct) eigenvalues of T, then V = V T (λ 1 ) V T (λ k ). (GED) (as V T (λ) = 0 whenever λ is not an eigenvalue of T ). In other words, every v V can be expressed uniquely as a sum v = v v k with v i V T (λ i ). Moreover, there is a basis of V with respect to which the matrix of T is block-diagonal of the form B 1 λ i..., where B i = (zeros below the diagonal) B k... λi and the size of the block B i is the eigenvalue multiplicity dim V T (λ i ). This theorem, which is also known as the primary decomposition theorem (in the case when F is algebraically closed), is basic to all that follows. The proof will come after we establish a few basic facts about generalized eigenvectors. 8. Basic facts about generalized eigenvectors. (a) Generalized eigenvectors v 1,..., v k corresponding, respectively, to distinct eigenvalues λ 1,..., λ k, are linearly independent. Proof. The identical technique in 3 applies here, too, but for slightly more complicated operators S (in fact, the complication is essentially notational). Suppose a 1 v 1 + a k v k = 0 with a i F. Let N be a positive integer that is so large that (T λ i ) N v i = 0 for i = 1,..., k, and let m be the smallest non-negative integer such that (T λ 1 ) m+1 x 1 = 0, i.e., (T λ 1 ) m x 1 0 is a λ 1 -eigenvector. To show that a 1 = 0, we let S = (T λ 1 ) m (T λ 2 ) N (T λ k ) N. We have (as in 3), Sx 2 =... = Sx k = 0. To compute Sx 1, we should express S in a form that distinguishes the nilpotency ( 15) of (T λ 1 ) on x 1. Thus, we write S = (T λ 1 ) m i 1( (T λ1 ) + (λ 1 λ i ) ) N. Page 3 of 9

4 In this form, we can see that Sx 1 = i 1 (λ 1 λ i ) N x 1 0, as follows. Expanding each factor ( (T λ 1 ) + (λ 1 λ i ) ) N according to the binomial theorem, we see that the whole product S acting on x 1 evalutes to a sum of the form Sx 1 = i 1 (λ 1 λ i ) N x 1 + terms of the form (scalar) (T λ 1 ) l x 1, with l > m, and each term of the second form is zero, because already (T λ 1 ) m+1 x 1 = 0. Now, applying S to the zero-sum a 1 v 1 + +a k v k gives 0 = a 1 Sx 1 0, which forces a 1 to be zero. Cleary, the same technique can be used to show that the remaining coeffcients are also zero. (b) V T (λ) = ker{(t λ) n }, where n = dim V. Proof. We assume that λ is an eigenvalue of T, for otherwise there s nothing to prove (both sides are zero). Since V is finite-dimensional, there is certainly a positive integer k such that V T (λ) = ker{(t λ) k }; let k be the smallest such integer. Clearly, it will suffice to show that k n. [To see that such a k exists, observe that we have an increasing chain of subspaces of V T (λ), ker(t λ) ker(t λ) 2 ker(t λ) 3, which must eventually stabilize, since dim V T (λ) <.] To this end, we pick a nonzero x V T (λ) with the property (T λ) k 1 x 0. We claim that the k vectors x, (T λ)x,..., (T λ) k 1 x are linearly independent, which would imply that k n. The strategy is of course! the one of part (a): Thus, we suppose a 0 x + a 1 (T λ)x + + a k 1 (T λ) k 1 = 0 and apply the operator S = (T λ) k 1 to the zero-sum, and observe that all the terms are immediately annihilated, leaving us with a 0 (T λ) k 1 x = 0. Since (T λ) k 1 x 0, we get a 0 = 0. Again, the same technique will give a 1 = 0, a 2 = 0, etc. (c) For each λ F, the generalized eigenspace V T (λ) is T -invariant, i.e., if x V T (λ), then T x V T (λ). Moreover, if λ is an eigenvalue of T, then λ is the only eigenvalue of the restriction of T to V T (λ). Proof. T -invariance is clear, because T commutes with any power of T λ. To demonstrate the other statement, let λ be an eigenvalue of the restriction of T to V T (λ), and let x be a corresponding eigenvector. From part (b), we have 0 = (T λ) n x = (λ λ) n x. Since x 0, we must therefore have λ = λ. 9. Proof of the generalized eigenspace decomposition ( 7). Let λ be an eigenvalue of T, and let n = dim V. We claim that 7(GED) will follow once we show that V = V T (λ) Im(T λ) n. (2) Indeed, suppose this direct-sum decomposition holds. Then the image W = Im(T λ) n is T -invariant, and therefore we can reason by induction on the dimension of the space, that W has a decomposition according to (GED) (observing, additionally, that (GED) is trivially true in the 1-dimensional case). Consequently, (GED) for the lower dimensional subspace W will yield (GED) for V. First notice that, by the dimension formula, V T (λ) = ker(t λ) n and Im(T λ) n will span V so long as V T (λ) Im(T λ) n = 0. So suppose x V T (λ) Im(T λ) n. Then Page 4 of 9

5 x = (T λ) n y for some y V. By 8(b), we have 0 = (T λ) n x = (T λ) 2n y, which implies that y V T (λ) = ker(t λ) n, and hence x = (T λ) n y = 0. This proves Eq. (2), and the decomposition 7(GED) now follows by applying the (appropriate) inductive hypothesis to W. [Exercise: Formulate this hypothesis explicitly!] Now we show the existence of a basis with the asserted properties. Since T maps each generalized eigenspace to itself ( 8(c)), and V is the direct sum of its generalized eigenspaces, any basis of V made up of generalized eigenvectors will express T as a block-diagonal matrix. So, we only need to show that each generalized eigenspace V T (λ i ) has a basis for which the matrix of T (restricted to V T (λ i )) is upper-triangular with λ i s along the diagonal. To this end, we examine the operator N = T λ i regarded as an operator on V T (λ i ). From 8(b) we know that N is nilpotent ( 15). Hence one of the subspaces in the the chain of subspaces ker N ker N 2 ker N 3 must eventually equal V T (λ i ). Thus, if we start by picking a basis of ker N, and extend that to a basis of ker N 2, and extend that to a basis of ker N 3, etc., then by this eventually terminating process, we will get a basis of V T (λ i ). With respect to this basis, the matrix of N is clearly strictly upper-triangular, and so the matrix of T = λ i I + N is upper-triangular with λ i s along the diagonal. 10. The minimal polynomial of an operator. Since the F -vector space End(V ) of linear operators is finite-dimensional, there is a smallest non-negative integer d such that the operators I, T, T 2,..., T d (I = identity operator) are linearly dependent (as elements of End(V )). Hence there are elements a i F such that The polynomial 0 = a 0 + a 1 T + + a d 1 T d 1 + T d m T (X) = a 0 + a 1 X + + a d 1 X d 1 + X d F [X] is called the minimal polynomial of T. It has the property that it is the monic polynomial p(x) (i.e., polynomial with highest-power coefficient 1) of smallest degree such that p(t ) = Factors of the minimal polynomial. Suppose F is algebraically closed. Let λ 1,..., λ k be the eigenvalues of T, and let ν i be the smallest positive integer such that V T (λ i ) = ker(t λ i ) νi. Then m T (X) = (X λ 1 ) ν1 (X λ k ) ν k. (3) Consequently, by 7(GED) and 8(b), the degree of m T (X) is no more than dim V. Moreover, m T (X) divides any polynomial p(x) F [X] such that p(t ) = 0. Proof. We first show that (X λ i ) νi divides m T (X). Let x 0 be a λ i -eigenvector of T. Then 0 = m T (T )x = m T (λ i )x, which implies that m T (λ i ) = 0. Hence there is a largest integer µ i such that (X λ i ) µi divides m T (X). To show that µ i ν i, we assume the contrary and derive a contradiction. If µ i < ν i, then there exists some x V T (λ i ) such that (T λ i ) µi x 0 (this follows from the very characterization of ν i ). Let q(x) be the monic polynomial m T (X)/(X λ i ) µi. Because F is algebraically closed, we can factor q(x), say q(x) = (X λ 1) (X λ l ). Then 0 = m T (T )x = q(t )(T λ i ) µi x. Now we argue as in 2 (second proof): Since y = (T λ i ) µi x 0 and 0 = q(t )y = (T λ 1) (T λ l )y, one of Page 5 of 9

6 the factors T λ j must annihilate a nonzero vector in V T (λ i ) (a vector in V T (λ i ), because y V T (λ i ), and V T (λ i ) is stable under T ). But the restriction of T to V T (λ i ) has only one eigenvalue, namely λ i ( 8(c)). Hence λ i = λ j. But this is absurd, for λ j is a zero of q(x), while by the very definition of q(x), λ i is not a zero of q(x). We conclude that µ i ν i. Now we eliminate the possibility that µ i > ν i. Indeed, from 7(GED) and the characterizing property of the ν i s, we know that for the polynomial m(x) = (X λ 1 ) ν1 (X λ k ) ν k, we have m(t ) = 0. Since µ i ν i, m(x) divides m T (X). But m T (X), by definition, is the lowest degree polynomial annihilating T. Hence not only must m(x) divide m T (X), it must equal m T (X). We have now established the factorization (3). As for the last assertion of the theorem, notice that in the first paragraph of our proof, we, in fact, showed that (X λ 1 ) ν1 (X λ k ) ν k divides p(t ) whenever p(t ) = 0. (The minimality of m T (X) was only used in the second paragraph.) 12. The characteristic polynomial. The polynomial c T (X) = det(xi T ) is called the characteristic polynomial of T. It is a monic polynomial of degree dim V. As we showed in 1, λ F is an eigenvalue of T if and only if c T (λ) = 0. Thus, if F is algebraically closed, then we have a factorization c T (X) = (X λ 1 ) µ1 (X λ k ) µ k, where the λ i s are the eigenvalues of T. The exponents µ i are, in fact, the eigenvalue multiplicities ( 6), i.e., µ i = dim V T (λ i ). This is an easy consequence of the equality c T (X) = det(xi T ) = det(xi A), where A is the block-diagonal matrix appearing in the statement of the generalized eigenspace decomposition ( 7). The fact that T is also a zero of its characterstic polynomial is the content of the well-known Cayley-Hamilton theorem. 13. The Cayley-Hamilton theorem. Let c T (X) = det(xi T ) be the characteristic polynomial of T. Then c T (T ) = 0. Proof. Since m T (T ) = 0, (recall that m T (X) is the minimal polynomial of T, see 10), it suffices to show that m T (X) divides c T (X). The theorem is valid over an arbitrary field, but we will only prove it under the simplifying assumption that F is algebraically closed. The minimal polynomial of T then equals m T (X) = (X λ 1 ) ν1 (X λ k ) ν k, where ν i is the smallest integer such that V T (λ i ) = ker(t λ i ) νi (see 11). From 8(b) it follows that ν i dim V T (λ i ) = µ i. Hence m T (X) divides c T (X) = (X λ 1 ) µ1 (X λ k ) µ k. 14. The correct interpretation of the Cayley-Hamilton theorem. It is obvious that det(t I T ) = 0: but be careful this triviality is not the content of the Cayley-Hamilton theorem! Rather, one needs first to compute the polynomial det(xi T ), and then substitute T for X. In terms of matrices, the characteristic polynomial c T (X) = det(xi T ) is the determinant of a matrix with entries in the ring F [X], not in the field F. For example, if T = ( ) , then the Cayley-Hamilton theorem is the assertion that T 2 3T + 2I = 0. Page 6 of 9

7 15. Nilpotent operators. A linear operator N: V V is said to be nilpotent if N d = 0 for some d 1. The smallest power d such that N d = 0 is called the order of N. It is easy to check that 0 is the only eigenvalue of N. [Do it!] Therefore, by 11(3), the minimal polynomial of a nilpotent operator of order d is X d. Moreover, there is basis of V with respect to which N is (strictly) upper-triangular [Homework exercise! but see 9]. 16. Cyclic subspaces: Definition. A cyclic subspace of V, or more precisely, a T -cyclic subspace, is a subspace of V of the form F [T ]x := { f(t )x f(t ) is an polynomial in T with coefficients in F }, for some x V. Actually, this only defines F [T ]x as a set, but it s easy to check that we, indeed, get a subspace of V. We shall be particularly interested in the case where T is nilpotent. 17. Main theorem 2: Cyclic decomposition for a nilpotent operator. If N is a nilpotent operator of V, then there are vectors x 1,..., x k of V such that V = F [N]x 1 F [N]x k. Proof. Here s a more verbose account of the sketch I gave in class. Let d be the nilpotence degree of N, i.e., N d = 0, but N d 1 0. Pick a vector x 1 V such that N d 1 x 1 0, and consider the cyclic subspace Z 1 = F [N]x 1. Since Z 1 is N-stable (i.e., N(Z 1 ) Z 1 ), N descends to a nilpotent operator Ñ on the quotient V/Z 1. Precisely, this means that the operator Ñ: V/Z 1 V/Z 1 defined by Ñ(x + Z 1) = N(x) + Z 1 is a well-defined linear operator (i.e., if x x Z 1 then Ñ(x + Z 1) = Ñ(x + Z 1 )). Since dim(v/z 1 ) = dim V dim Z 1 < dim V, we may argue by induction on the dimension to conclude that V/Z 1 has a cyclic decomposition, say V/Z 1 = F [Ñ] x 2 F [Ñ] x m (noting also that the theorem is true, trivially, in the 1-dimensional case). The only tricky part of the proof (the main part, really) is to lift this decomposition to V, and show that it complements Z 1. To this end, it suffices to establish the following claim: If d i is the period of the Ñ-cycle { x i, Ñ x i, Ñ 2 x i,... } (which is to say, d i is the smallest integer such that Ñ di x i = 0), then x i has a representative x i (i.e., an element x i V such that x i = x i + Z 1 ) whose N-cycle also has period d i. and Assuming this, the proof now basically wraps itself up, for we have: V = Z 1 + F [N]x F [N]x k dim V = dim Z 1 + dim(v/z 1 ) = dim Z 1 + dim(f [N]x 2 ) + + dim(f [N]x k ) (since dim(f [N]x 2 ) = dim(f [Ñ] x 2), as the N-period of x i equals the Ñ-period of x i cf. Proof of 8(b)) which implies that V = Z 1 F [N]x 2 F [N]x k. Page 7 of 9

8 To establish the claim for x 2, say, start by picking any representative x of x 2. Since N d2 x Z 1, we have N d2 x = p(n)x 1 for some polynomial p(x). Thus 0 = N d x = N d d2 p(n)x 1. Since the N-cycle of x 1 has length d, X d2 must divide p(x), say p(x) = X d2 q(x). Let x 2 = x q(n)x 1. Then x 2 represents x 2 and N d2 x 2 = 0, and the claim follows (obviously, N s x 2 0 whenever s < d 2 ). Clearly, the same argument applies to establish the analogous claim for the remaining cyclic generators x 3,..., x k. The theorem is thereby proved. 18. Jordan matrices: Terminology. It is important to be clear here about what we mean by Jordan matrix, because there are slight variations of usage in the literature, and by having names for various kinds of Jordan matrices we can also avoid having to write big matrices everywhere. Let λ F. A simple Jordan (or λ-jordan) matrix of size k is the k k matrix J k (λ) = λ 1 λ λ 1 λ (blanks are zero). Observe that a simple Jordan matrix J k (λ) can be expressed as a sum of a scalar matrix (namely, λi) plus a nilpotent matrix of order k (namely, J k (λ) λi, which is the matrix with 1 s just above the main diagonal and zeros elsewhere). Observe also that λ is the unique eigenvalue of J k (λ), and that space of its eigenvectors is 1-dimensional. A Jordan (or λ-jordan) matrix is any block-diagonal matrix of the form J l (λ) J m (λ).... Jn(λ) 19. Jordan canonical form of a nilpotent operator. If N is a nilpotent operator of V, then there is a basis of V with respect to which the matrix of N is a 0-Jordan matrix. Proof. By the cyclic decomposition for a nilpotent operator ( 17), there are vectors x 1,..., x k of V such that V = F [N]x 1 F [N]x k. For each x i, let α i be the smallest positive integer such that N αi x i = 0. From the proof of 8(b), it follows that the vectors N α1 1 x 1,..., Nx 1, x 1,..., N α k 1 x k,..., Nx k, x k, form an ordered basis for V. With respect to this basis, the matrix of N is a 0-Jordan matrix with k blocks of sizes α 1,..., α k. 20. Jordan canonical form, in general. Suppose T : V V is a linear transformation all of whose (distinct) eigenvalues, say λ 1,..., λ m, lie in F (i.e., the characteristic/minimal polynomial of T is a product of linear factors with coefficients in F ). Then there is a basis of V with respect to which the matrix of T takes the block-diagonal form J = J λ1 J λ2... Jλm Page 8 of 9

9 where each J λi is an λ i -Jordan matrix. Moreover, for each eigenvalue λ of T, if β k (λ) denotes the number of elementary λ-jordan matrices of size k (within J λ ), then β k (λ) = rank{(t λ) k 1 } 2 rank{(t λ) k } + rank{(t λ) k+1 }. (4) In particular, the matrix J is uniquely determined by T up to permutation of its Jordan blocks, because formula (4) only depends on T. You should check that formula (4) gives zero when λ is not an eigenvalue of T (as it should). We call J the Jordan canonical form of T. 21. Computing the Jordan canonical form. Because we have an efficient means of computing the rank of a transformation (method of row reduction), the rank formula 20(4) gives an efficient way to compute the Jordan canonical form (when the eigenvalues of T are known). You will get homework exercise working out the Jordan canonical form concretely. Additionally, consult Hoffman-Kunze (or any other linear algebra textbook) for other worked examples. 22. Proof of the existence of the Jordan canonical form. Since T maps each generalized eigenspace to itself ( 8(c)), the generalized eigenspace decomposition of V ( 7) shows that it suffices to prove that each generalized eigenspace V T (λ) (for λ an eigenvalue of T ) has a basis with respect to which the matrix of T (or rather, of the restriction of T to V T (λ)) is a λ-jordan matrix. By 8(b), the operator (T λ) is nilpotent on V T (λ). By 19, we know that V T (λ) then has a basis with respect to which the matrix of (T λ) is a 0-Jordan matrix; equivalently, the matrix of T = (T λ) + λ in this basis is a λ-jordan matrix. Verification of the rank formula (4) is left as a homework exercise (with copious hints)! Page 9 of 9

ALGEBRA QUALIFYING EXAM PROBLEMS LINEAR ALGEBRA

ALGEBRA QUALIFYING EXAM PROBLEMS LINEAR ALGEBRA ALGEBRA QUALIFYING EXAM PROBLEMS LINEAR ALGEBRA Kent State University Department of Mathematical Sciences Compiled and Maintained by Donald L. White Version: August 29, 2017 CONTENTS LINEAR ALGEBRA AND

More information

The Jordan Canonical Form

The Jordan Canonical Form The Jordan Canonical Form The Jordan canonical form describes the structure of an arbitrary linear transformation on a finite-dimensional vector space over an algebraically closed field. Here we develop

More information

Topics in linear algebra

Topics in linear algebra Chapter 6 Topics in linear algebra 6.1 Change of basis I want to remind you of one of the basic ideas in linear algebra: change of basis. Let F be a field, V and W be finite dimensional vector spaces over

More information

LINEAR ALGEBRA BOOT CAMP WEEK 2: LINEAR OPERATORS

LINEAR ALGEBRA BOOT CAMP WEEK 2: LINEAR OPERATORS LINEAR ALGEBRA BOOT CAMP WEEK 2: LINEAR OPERATORS Unless otherwise stated, all vector spaces in this worksheet are finite dimensional and the scalar field F has characteristic zero. The following are facts

More information

GENERALIZED EIGENVECTORS, MINIMAL POLYNOMIALS AND THEOREM OF CAYLEY-HAMILTION

GENERALIZED EIGENVECTORS, MINIMAL POLYNOMIALS AND THEOREM OF CAYLEY-HAMILTION GENERALIZED EIGENVECTORS, MINIMAL POLYNOMIALS AND THEOREM OF CAYLEY-HAMILTION FRANZ LUEF Abstract. Our exposition is inspired by S. Axler s approach to linear algebra and follows largely his exposition

More information

LINEAR ALGEBRA BOOT CAMP WEEK 1: THE BASICS

LINEAR ALGEBRA BOOT CAMP WEEK 1: THE BASICS LINEAR ALGEBRA BOOT CAMP WEEK 1: THE BASICS Unless otherwise stated, all vector spaces in this worksheet are finite dimensional and the scalar field F has characteristic zero. The following are facts (in

More information

NONCOMMUTATIVE POLYNOMIAL EQUATIONS. Edward S. Letzter. Introduction

NONCOMMUTATIVE POLYNOMIAL EQUATIONS. Edward S. Letzter. Introduction NONCOMMUTATIVE POLYNOMIAL EQUATIONS Edward S Letzter Introduction My aim in these notes is twofold: First, to briefly review some linear algebra Second, to provide you with some new tools and techniques

More information

Linear Algebra Final Exam Solutions, December 13, 2008

Linear Algebra Final Exam Solutions, December 13, 2008 Linear Algebra Final Exam Solutions, December 13, 2008 Write clearly, with complete sentences, explaining your work. You will be graded on clarity, style, and brevity. If you add false statements to a

More information

1 Invariant subspaces

1 Invariant subspaces MATH 2040 Linear Algebra II Lecture Notes by Martin Li Lecture 8 Eigenvalues, eigenvectors and invariant subspaces 1 In previous lectures we have studied linear maps T : V W from a vector space V to another

More information

Definition (T -invariant subspace) Example. Example

Definition (T -invariant subspace) Example. Example Eigenvalues, Eigenvectors, Similarity, and Diagonalization We now turn our attention to linear transformations of the form T : V V. To better understand the effect of T on the vector space V, we begin

More information

THE MINIMAL POLYNOMIAL AND SOME APPLICATIONS

THE MINIMAL POLYNOMIAL AND SOME APPLICATIONS THE MINIMAL POLYNOMIAL AND SOME APPLICATIONS KEITH CONRAD. Introduction The easiest matrices to compute with are the diagonal ones. The sum and product of diagonal matrices can be computed componentwise

More information

JORDAN NORMAL FORM. Contents Introduction 1 Jordan Normal Form 1 Conclusion 5 References 5

JORDAN NORMAL FORM. Contents Introduction 1 Jordan Normal Form 1 Conclusion 5 References 5 JORDAN NORMAL FORM KATAYUN KAMDIN Abstract. This paper outlines a proof of the Jordan Normal Form Theorem. First we show that a complex, finite dimensional vector space can be decomposed into a direct

More information

EXERCISES AND SOLUTIONS IN LINEAR ALGEBRA

EXERCISES AND SOLUTIONS IN LINEAR ALGEBRA EXERCISES AND SOLUTIONS IN LINEAR ALGEBRA Mahmut Kuzucuoğlu Middle East Technical University matmah@metu.edu.tr Ankara, TURKEY March 14, 015 ii TABLE OF CONTENTS CHAPTERS 0. PREFACE..................................................

More information

Infinite-Dimensional Triangularization

Infinite-Dimensional Triangularization Infinite-Dimensional Triangularization Zachary Mesyan March 11, 2018 Abstract The goal of this paper is to generalize the theory of triangularizing matrices to linear transformations of an arbitrary vector

More information

Homework For each of the following matrices, find the minimal polynomial and determine whether the matrix is diagonalizable.

Homework For each of the following matrices, find the minimal polynomial and determine whether the matrix is diagonalizable. Math 5327 Fall 2018 Homework 7 1. For each of the following matrices, find the minimal polynomial and determine whether the matrix is diagonalizable. 3 1 0 (a) A = 1 2 0 1 1 0 x 3 1 0 Solution: 1 x 2 0

More information

The Cayley-Hamilton Theorem and the Jordan Decomposition

The Cayley-Hamilton Theorem and the Jordan Decomposition LECTURE 19 The Cayley-Hamilton Theorem and the Jordan Decomposition Let me begin by summarizing the main results of the last lecture Suppose T is a endomorphism of a vector space V Then T has a minimal

More information

Math 113 Midterm Exam Solutions

Math 113 Midterm Exam Solutions Math 113 Midterm Exam Solutions Held Thursday, May 7, 2013, 7-9 pm. 1. (10 points) Let V be a vector space over F and T : V V be a linear operator. Suppose that there is a non-zero vector v V such that

More information

A linear algebra proof of the fundamental theorem of algebra

A linear algebra proof of the fundamental theorem of algebra A linear algebra proof of the fundamental theorem of algebra Andrés E. Caicedo May 18, 2010 Abstract We present a recent proof due to Harm Derksen, that any linear operator in a complex finite dimensional

More information

A linear algebra proof of the fundamental theorem of algebra

A linear algebra proof of the fundamental theorem of algebra A linear algebra proof of the fundamental theorem of algebra Andrés E. Caicedo May 18, 2010 Abstract We present a recent proof due to Harm Derksen, that any linear operator in a complex finite dimensional

More information

Math 113 Final Exam: Solutions

Math 113 Final Exam: Solutions Math 113 Final Exam: Solutions Thursday, June 11, 2013, 3.30-6.30pm. 1. (25 points total) Let P 2 (R) denote the real vector space of polynomials of degree 2. Consider the following inner product on P

More information

Math 113 Homework 5. Bowei Liu, Chao Li. Fall 2013

Math 113 Homework 5. Bowei Liu, Chao Li. Fall 2013 Math 113 Homework 5 Bowei Liu, Chao Li Fall 2013 This homework is due Thursday November 7th at the start of class. Remember to write clearly, and justify your solutions. Please make sure to put your name

More information

Math 594. Solutions 5

Math 594. Solutions 5 Math 594. Solutions 5 Book problems 6.1: 7. Prove that subgroups and quotient groups of nilpotent groups are nilpotent (your proof should work for infinite groups). Give an example of a group G which possesses

More information

MATHEMATICS 217 NOTES

MATHEMATICS 217 NOTES MATHEMATICS 27 NOTES PART I THE JORDAN CANONICAL FORM The characteristic polynomial of an n n matrix A is the polynomial χ A (λ) = det(λi A), a monic polynomial of degree n; a monic polynomial in the variable

More information

OHSx XM511 Linear Algebra: Solutions to Online True/False Exercises

OHSx XM511 Linear Algebra: Solutions to Online True/False Exercises This document gives the solutions to all of the online exercises for OHSx XM511. The section ( ) numbers refer to the textbook. TYPE I are True/False. Answers are in square brackets [. Lecture 02 ( 1.1)

More information

Math 110 Linear Algebra Midterm 2 Review October 28, 2017

Math 110 Linear Algebra Midterm 2 Review October 28, 2017 Math 11 Linear Algebra Midterm Review October 8, 17 Material Material covered on the midterm includes: All lectures from Thursday, Sept. 1st to Tuesday, Oct. 4th Homeworks 9 to 17 Quizzes 5 to 9 Sections

More information

Given a finite-dimensional vector space V over a field K, recall that a linear

Given a finite-dimensional vector space V over a field K, recall that a linear Jordan normal form Sebastian Ørsted December 16, 217 Abstract In these notes, we expand upon the coverage of linear algebra as presented in Thomsen (216). Namely, we introduce some concepts and results

More information

0.1 Rational Canonical Forms

0.1 Rational Canonical Forms We have already seen that it is useful and simpler to study linear systems using matrices. But matrices are themselves cumbersome, as they are stuffed with many entries, and it turns out that it s best

More information

The Cyclic Decomposition of a Nilpotent Operator

The Cyclic Decomposition of a Nilpotent Operator The Cyclic Decomposition of a Nilpotent Operator 1 Introduction. J.H. Shapiro Suppose T is a linear transformation on a vector space V. Recall Exercise #3 of Chapter 8 of our text, which we restate here

More information

Honors Algebra 4, MATH 371 Winter 2010 Assignment 4 Due Wednesday, February 17 at 08:35

Honors Algebra 4, MATH 371 Winter 2010 Assignment 4 Due Wednesday, February 17 at 08:35 Honors Algebra 4, MATH 371 Winter 2010 Assignment 4 Due Wednesday, February 17 at 08:35 1. Let R be a commutative ring with 1 0. (a) Prove that the nilradical of R is equal to the intersection of the prime

More information

A proof of the Jordan normal form theorem

A proof of the Jordan normal form theorem A proof of the Jordan normal form theorem Jordan normal form theorem states that any matrix is similar to a blockdiagonal matrix with Jordan blocks on the diagonal. To prove it, we first reformulate it

More information

Solution. That ϕ W is a linear map W W follows from the definition of subspace. The map ϕ is ϕ(v + W ) = ϕ(v) + W, which is well-defined since

Solution. That ϕ W is a linear map W W follows from the definition of subspace. The map ϕ is ϕ(v + W ) = ϕ(v) + W, which is well-defined since MAS 5312 Section 2779 Introduction to Algebra 2 Solutions to Selected Problems, Chapters 11 13 11.2.9 Given a linear ϕ : V V such that ϕ(w ) W, show ϕ induces linear ϕ W : W W and ϕ : V/W V/W : Solution.

More information

(VI.D) Generalized Eigenspaces

(VI.D) Generalized Eigenspaces (VI.D) Generalized Eigenspaces Let T : C n C n be a f ixed linear transformation. For this section and the next, all vector spaces are assumed to be over C ; in particular, we will often write V for C

More information

Lecture 21: The decomposition theorem into generalized eigenspaces; multiplicity of eigenvalues and upper-triangular matrices (1)

Lecture 21: The decomposition theorem into generalized eigenspaces; multiplicity of eigenvalues and upper-triangular matrices (1) Lecture 21: The decomposition theorem into generalized eigenspaces; multiplicity of eigenvalues and upper-triangular matrices (1) Travis Schedler Tue, Nov 29, 2011 (version: Tue, Nov 29, 1:00 PM) Goals

More information

Theorem 5.3. Let E/F, E = F (u), be a simple field extension. Then u is algebraic if and only if E/F is finite. In this case, [E : F ] = deg f u.

Theorem 5.3. Let E/F, E = F (u), be a simple field extension. Then u is algebraic if and only if E/F is finite. In this case, [E : F ] = deg f u. 5. Fields 5.1. Field extensions. Let F E be a subfield of the field E. We also describe this situation by saying that E is an extension field of F, and we write E/F to express this fact. If E/F is a field

More information

NOTES II FOR 130A JACOB STERBENZ

NOTES II FOR 130A JACOB STERBENZ NOTES II FOR 130A JACOB STERBENZ Abstract. Here are some notes on the Jordan canonical form as it was covered in class. Contents 1. Polynomials 1 2. The Minimal Polynomial and the Primary Decomposition

More information

4.1 Eigenvalues, Eigenvectors, and The Characteristic Polynomial

4.1 Eigenvalues, Eigenvectors, and The Characteristic Polynomial Linear Algebra (part 4): Eigenvalues, Diagonalization, and the Jordan Form (by Evan Dummit, 27, v ) Contents 4 Eigenvalues, Diagonalization, and the Jordan Canonical Form 4 Eigenvalues, Eigenvectors, and

More information

MATH SOLUTIONS TO PRACTICE MIDTERM LECTURE 1, SUMMER Given vector spaces V and W, V W is the vector space given by

MATH SOLUTIONS TO PRACTICE MIDTERM LECTURE 1, SUMMER Given vector spaces V and W, V W is the vector space given by MATH 110 - SOLUTIONS TO PRACTICE MIDTERM LECTURE 1, SUMMER 2009 GSI: SANTIAGO CAÑEZ 1. Given vector spaces V and W, V W is the vector space given by V W = {(v, w) v V and w W }, with addition and scalar

More information

Ir O D = D = ( ) Section 2.6 Example 1. (Bottom of page 119) dim(v ) = dim(l(v, W )) = dim(v ) dim(f ) = dim(v )

Ir O D = D = ( ) Section 2.6 Example 1. (Bottom of page 119) dim(v ) = dim(l(v, W )) = dim(v ) dim(f ) = dim(v ) Section 3.2 Theorem 3.6. Let A be an m n matrix of rank r. Then r m, r n, and, by means of a finite number of elementary row and column operations, A can be transformed into the matrix ( ) Ir O D = 1 O

More information

Homework 6 Solutions. Solution. Note {e t, te t, t 2 e t, e 2t } is linearly independent. If β = {e t, te t, t 2 e t, e 2t }, then

Homework 6 Solutions. Solution. Note {e t, te t, t 2 e t, e 2t } is linearly independent. If β = {e t, te t, t 2 e t, e 2t }, then Homework 6 Solutions 1 Let V be the real vector space spanned by the functions e t, te t, t 2 e t, e 2t Find a Jordan canonical basis and a Jordan canonical form of T on V dened by T (f) = f Solution Note

More information

Remark By definition, an eigenvector must be a nonzero vector, but eigenvalue could be zero.

Remark By definition, an eigenvector must be a nonzero vector, but eigenvalue could be zero. Sec 6 Eigenvalues and Eigenvectors Definition An eigenvector of an n n matrix A is a nonzero vector x such that A x λ x for some scalar λ A scalar λ is called an eigenvalue of A if there is a nontrivial

More information

Generalized eigenspaces

Generalized eigenspaces Generalized eigenspaces November 30, 2012 Contents 1 Introduction 1 2 Polynomials 2 3 Calculating the characteristic polynomial 5 4 Projections 7 5 Generalized eigenvalues 10 6 Eigenpolynomials 15 1 Introduction

More information

1.4 Solvable Lie algebras

1.4 Solvable Lie algebras 1.4. SOLVABLE LIE ALGEBRAS 17 1.4 Solvable Lie algebras 1.4.1 Derived series and solvable Lie algebras The derived series of a Lie algebra L is given by: L (0) = L, L (1) = [L, L],, L (2) = [L (1), L (1)

More information

MATH 1120 (LINEAR ALGEBRA 1), FINAL EXAM FALL 2011 SOLUTIONS TO PRACTICE VERSION

MATH 1120 (LINEAR ALGEBRA 1), FINAL EXAM FALL 2011 SOLUTIONS TO PRACTICE VERSION MATH (LINEAR ALGEBRA ) FINAL EXAM FALL SOLUTIONS TO PRACTICE VERSION Problem (a) For each matrix below (i) find a basis for its column space (ii) find a basis for its row space (iii) determine whether

More information

GRE Subject test preparation Spring 2016 Topic: Abstract Algebra, Linear Algebra, Number Theory.

GRE Subject test preparation Spring 2016 Topic: Abstract Algebra, Linear Algebra, Number Theory. GRE Subject test preparation Spring 2016 Topic: Abstract Algebra, Linear Algebra, Number Theory. Linear Algebra Standard matrix manipulation to compute the kernel, intersection of subspaces, column spaces,

More information

Math 396. Quotient spaces

Math 396. Quotient spaces Math 396. Quotient spaces. Definition Let F be a field, V a vector space over F and W V a subspace of V. For v, v V, we say that v v mod W if and only if v v W. One can readily verify that with this definition

More information

Linear Algebra 1. M.T.Nair Department of Mathematics, IIT Madras. and in that case x is called an eigenvector of T corresponding to the eigenvalue λ.

Linear Algebra 1. M.T.Nair Department of Mathematics, IIT Madras. and in that case x is called an eigenvector of T corresponding to the eigenvalue λ. Linear Algebra 1 M.T.Nair Department of Mathematics, IIT Madras 1 Eigenvalues and Eigenvectors 1.1 Definition and Examples Definition 1.1. Let V be a vector space (over a field F) and T : V V be a linear

More information

Review problems for MA 54, Fall 2004.

Review problems for MA 54, Fall 2004. Review problems for MA 54, Fall 2004. Below are the review problems for the final. They are mostly homework problems, or very similar. If you are comfortable doing these problems, you should be fine on

More information

The converse is clear, since

The converse is clear, since 14. The minimal polynomial For an example of a matrix which cannot be diagonalised, consider the matrix ( ) 0 1 A =. 0 0 The characteristic polynomial is λ 2 = 0 so that the only eigenvalue is λ = 0. The

More information

(f + g)(s) = f(s) + g(s) for f, g V, s S (cf)(s) = cf(s) for c F, f V, s S

(f + g)(s) = f(s) + g(s) for f, g V, s S (cf)(s) = cf(s) for c F, f V, s S 1 Vector spaces 1.1 Definition (Vector space) Let V be a set with a binary operation +, F a field, and (c, v) cv be a mapping from F V into V. Then V is called a vector space over F (or a linear space

More information

Math 554 Qualifying Exam. You may use any theorems from the textbook. Any other claims must be proved in details.

Math 554 Qualifying Exam. You may use any theorems from the textbook. Any other claims must be proved in details. Math 554 Qualifying Exam January, 2019 You may use any theorems from the textbook. Any other claims must be proved in details. 1. Let F be a field and m and n be positive integers. Prove the following.

More information

j=1 x j p, if 1 p <, x i ξ : x i < ξ} 0 as p.

j=1 x j p, if 1 p <, x i ξ : x i < ξ} 0 as p. LINEAR ALGEBRA Fall 203 The final exam Almost all of the problems solved Exercise Let (V, ) be a normed vector space. Prove x y x y for all x, y V. Everybody knows how to do this! Exercise 2 If V is a

More information

Math 4153 Exam 3 Review. The syllabus for Exam 3 is Chapter 6 (pages ), Chapter 7 through page 137, and Chapter 8 through page 182 in Axler.

Math 4153 Exam 3 Review. The syllabus for Exam 3 is Chapter 6 (pages ), Chapter 7 through page 137, and Chapter 8 through page 182 in Axler. Math 453 Exam 3 Review The syllabus for Exam 3 is Chapter 6 (pages -2), Chapter 7 through page 37, and Chapter 8 through page 82 in Axler.. You should be sure to know precise definition of the terms we

More information

Notes on the matrix exponential

Notes on the matrix exponential Notes on the matrix exponential Erik Wahlén erik.wahlen@math.lu.se February 14, 212 1 Introduction The purpose of these notes is to describe how one can compute the matrix exponential e A when A is not

More information

MATH 583A REVIEW SESSION #1

MATH 583A REVIEW SESSION #1 MATH 583A REVIEW SESSION #1 BOJAN DURICKOVIC 1. Vector Spaces Very quick review of the basic linear algebra concepts (see any linear algebra textbook): (finite dimensional) vector space (or linear space),

More information

MATH 115A: SAMPLE FINAL SOLUTIONS

MATH 115A: SAMPLE FINAL SOLUTIONS MATH A: SAMPLE FINAL SOLUTIONS JOE HUGHES. Let V be the set of all functions f : R R such that f( x) = f(x) for all x R. Show that V is a vector space over R under the usual addition and scalar multiplication

More information

Study Guide for Linear Algebra Exam 2

Study Guide for Linear Algebra Exam 2 Study Guide for Linear Algebra Exam 2 Term Vector Space Definition A Vector Space is a nonempty set V of objects, on which are defined two operations, called addition and multiplication by scalars (real

More information

SPRING 2006 PRELIMINARY EXAMINATION SOLUTIONS

SPRING 2006 PRELIMINARY EXAMINATION SOLUTIONS SPRING 006 PRELIMINARY EXAMINATION SOLUTIONS 1A. Let G be the subgroup of the free abelian group Z 4 consisting of all integer vectors (x, y, z, w) such that x + 3y + 5z + 7w = 0. (a) Determine a linearly

More information

Algebra Homework, Edition 2 9 September 2010

Algebra Homework, Edition 2 9 September 2010 Algebra Homework, Edition 2 9 September 2010 Problem 6. (1) Let I and J be ideals of a commutative ring R with I + J = R. Prove that IJ = I J. (2) Let I, J, and K be ideals of a principal ideal domain.

More information

Chapters 5 & 6: Theory Review: Solutions Math 308 F Spring 2015

Chapters 5 & 6: Theory Review: Solutions Math 308 F Spring 2015 Chapters 5 & 6: Theory Review: Solutions Math 308 F Spring 205. If A is a 3 3 triangular matrix, explain why det(a) is equal to the product of entries on the diagonal. If A is a lower triangular or diagonal

More information

Contents. Preface for the Instructor. Preface for the Student. xvii. Acknowledgments. 1 Vector Spaces 1 1.A R n and C n 2

Contents. Preface for the Instructor. Preface for the Student. xvii. Acknowledgments. 1 Vector Spaces 1 1.A R n and C n 2 Contents Preface for the Instructor xi Preface for the Student xv Acknowledgments xvii 1 Vector Spaces 1 1.A R n and C n 2 Complex Numbers 2 Lists 5 F n 6 Digression on Fields 10 Exercises 1.A 11 1.B Definition

More information

MATH 205 HOMEWORK #3 OFFICIAL SOLUTION. Problem 1: Find all eigenvalues and eigenvectors of the following linear transformations. (a) F = R, V = R 3,

MATH 205 HOMEWORK #3 OFFICIAL SOLUTION. Problem 1: Find all eigenvalues and eigenvectors of the following linear transformations. (a) F = R, V = R 3, MATH 205 HOMEWORK #3 OFFICIAL SOLUTION Problem 1: Find all eigenvalues and eigenvectors of the following linear transformations. a F = R, V = R 3, b F = R or C, V = F 2, T = T = 9 4 4 8 3 4 16 8 7 0 1

More information

Minimum Polynomials of Linear Transformations

Minimum Polynomials of Linear Transformations Minimum Polynomials of Linear Transformations Spencer De Chenne University of Puget Sound 30 April 2014 Table of Contents Polynomial Basics Endomorphisms Minimum Polynomial Building Linear Transformations

More information

Linear Algebra II Lecture 13

Linear Algebra II Lecture 13 Linear Algebra II Lecture 13 Xi Chen 1 1 University of Alberta November 14, 2014 Outline 1 2 If v is an eigenvector of T : V V corresponding to λ, then v is an eigenvector of T m corresponding to λ m since

More information

Jordan Normal Form. Chapter Minimal Polynomials

Jordan Normal Form. Chapter Minimal Polynomials Chapter 8 Jordan Normal Form 81 Minimal Polynomials Recall p A (x) =det(xi A) is called the characteristic polynomial of the matrix A Theorem 811 Let A M n Then there exists a unique monic polynomial q

More information

Chapter SSM: Linear Algebra. 5. Find all x such that A x = , so that x 1 = x 2 = 0.

Chapter SSM: Linear Algebra. 5. Find all x such that A x = , so that x 1 = x 2 = 0. Chapter Find all x such that A x : Chapter, so that x x ker(a) { } Find all x such that A x ; note that all x in R satisfy the equation, so that ker(a) R span( e, e ) 5 Find all x such that A x 5 ; x x

More information

Econ Slides from Lecture 7

Econ Slides from Lecture 7 Econ 205 Sobel Econ 205 - Slides from Lecture 7 Joel Sobel August 31, 2010 Linear Algebra: Main Theory A linear combination of a collection of vectors {x 1,..., x k } is a vector of the form k λ ix i for

More information

A connection between number theory and linear algebra

A connection between number theory and linear algebra A connection between number theory and linear algebra Mark Steinberger Contents 1. Some basics 1 2. Rational canonical form 2 3. Prime factorization in F[x] 4 4. Units and order 5 5. Finite fields 7 6.

More information

MATH 320: PRACTICE PROBLEMS FOR THE FINAL AND SOLUTIONS

MATH 320: PRACTICE PROBLEMS FOR THE FINAL AND SOLUTIONS MATH 320: PRACTICE PROBLEMS FOR THE FINAL AND SOLUTIONS There will be eight problems on the final. The following are sample problems. Problem 1. Let F be the vector space of all real valued functions on

More information

Remark 1 By definition, an eigenvector must be a nonzero vector, but eigenvalue could be zero.

Remark 1 By definition, an eigenvector must be a nonzero vector, but eigenvalue could be zero. Sec 5 Eigenvectors and Eigenvalues In this chapter, vector means column vector Definition An eigenvector of an n n matrix A is a nonzero vector x such that A x λ x for some scalar λ A scalar λ is called

More information

Math 110, Spring 2015: Midterm Solutions

Math 110, Spring 2015: Midterm Solutions Math 11, Spring 215: Midterm Solutions These are not intended as model answers ; in many cases far more explanation is provided than would be necessary to receive full credit. The goal here is to make

More information

Advanced Engineering Mathematics Prof. Pratima Panigrahi Department of Mathematics Indian Institute of Technology, Kharagpur

Advanced Engineering Mathematics Prof. Pratima Panigrahi Department of Mathematics Indian Institute of Technology, Kharagpur Advanced Engineering Mathematics Prof. Pratima Panigrahi Department of Mathematics Indian Institute of Technology, Kharagpur Lecture No. #07 Jordan Canonical Form Cayley Hamilton Theorem (Refer Slide Time:

More information

The Jordan canonical form

The Jordan canonical form The Jordan canonical form Francisco Javier Sayas University of Delaware November 22, 213 The contents of these notes have been translated and slightly modified from a previous version in Spanish. Part

More information

Further linear algebra. Chapter IV. Jordan normal form.

Further linear algebra. Chapter IV. Jordan normal form. Further linear algebra. Chapter IV. Jordan normal form. Andrei Yafaev In what follows V is a vector space of dimension n and B is a basis of V. In this chapter we are concerned with linear maps T : V V.

More information

LINEAR ALGEBRA 1, 2012-I PARTIAL EXAM 3 SOLUTIONS TO PRACTICE PROBLEMS

LINEAR ALGEBRA 1, 2012-I PARTIAL EXAM 3 SOLUTIONS TO PRACTICE PROBLEMS LINEAR ALGEBRA, -I PARTIAL EXAM SOLUTIONS TO PRACTICE PROBLEMS Problem (a) For each of the two matrices below, (i) determine whether it is diagonalizable, (ii) determine whether it is orthogonally diagonalizable,

More information

ALGEBRA QUALIFYING EXAM PROBLEMS

ALGEBRA QUALIFYING EXAM PROBLEMS ALGEBRA QUALIFYING EXAM PROBLEMS Kent State University Department of Mathematical Sciences Compiled and Maintained by Donald L. White Version: August 29, 2017 CONTENTS LINEAR ALGEBRA AND MODULES General

More information

MAT2342 : Introduction to Applied Linear Algebra Mike Newman, fall Projections. introduction

MAT2342 : Introduction to Applied Linear Algebra Mike Newman, fall Projections. introduction MAT4 : Introduction to Applied Linear Algebra Mike Newman fall 7 9. Projections introduction One reason to consider projections is to understand approximate solutions to linear systems. A common example

More information

CANONICAL FORMS FOR LINEAR TRANSFORMATIONS AND MATRICES. D. Katz

CANONICAL FORMS FOR LINEAR TRANSFORMATIONS AND MATRICES. D. Katz CANONICAL FORMS FOR LINEAR TRANSFORMATIONS AND MATRICES D. Katz The purpose of this note is to present the rational canonical form and Jordan canonical form theorems for my M790 class. Throughout, we fix

More information

MATH 315 Linear Algebra Homework #1 Assigned: August 20, 2018

MATH 315 Linear Algebra Homework #1 Assigned: August 20, 2018 Homework #1 Assigned: August 20, 2018 Review the following subjects involving systems of equations and matrices from Calculus II. Linear systems of equations Converting systems to matrix form Pivot entry

More information

University of Colorado at Denver Mathematics Department Applied Linear Algebra Preliminary Exam With Solutions 16 January 2009, 10:00 am 2:00 pm

University of Colorado at Denver Mathematics Department Applied Linear Algebra Preliminary Exam With Solutions 16 January 2009, 10:00 am 2:00 pm University of Colorado at Denver Mathematics Department Applied Linear Algebra Preliminary Exam With Solutions 16 January 2009, 10:00 am 2:00 pm Name: The proctor will let you read the following conditions

More information

Elementary linear algebra

Elementary linear algebra Chapter 1 Elementary linear algebra 1.1 Vector spaces Vector spaces owe their importance to the fact that so many models arising in the solutions of specific problems turn out to be vector spaces. The

More information

1 Linear transformations; the basics

1 Linear transformations; the basics Linear Algebra Fall 2013 Linear Transformations 1 Linear transformations; the basics Definition 1 Let V, W be vector spaces over the same field F. A linear transformation (also known as linear map, or

More information

A NOTE ON THE JORDAN CANONICAL FORM

A NOTE ON THE JORDAN CANONICAL FORM A NOTE ON THE JORDAN CANONICAL FORM H. Azad Department of Mathematics and Statistics King Fahd University of Petroleum & Minerals Dhahran, Saudi Arabia hassanaz@kfupm.edu.sa Abstract A proof of the Jordan

More information

MATH PRACTICE EXAM 1 SOLUTIONS

MATH PRACTICE EXAM 1 SOLUTIONS MATH 2359 PRACTICE EXAM SOLUTIONS SPRING 205 Throughout this exam, V and W will denote vector spaces over R Part I: True/False () For each of the following statements, determine whether the statement is

More information

The eigenvalues are the roots of the characteristic polynomial, det(a λi). We can compute

The eigenvalues are the roots of the characteristic polynomial, det(a λi). We can compute A. [ 3. Let A = 5 5 ]. Find all (complex) eigenvalues and eigenvectors of The eigenvalues are the roots of the characteristic polynomial, det(a λi). We can compute 3 λ A λi =, 5 5 λ from which det(a λi)

More information

Linear Algebra II Lecture 22

Linear Algebra II Lecture 22 Linear Algebra II Lecture 22 Xi Chen University of Alberta March 4, 24 Outline Characteristic Polynomial, Eigenvalue, Eigenvector and Eigenvalue, Eigenvector and Let T : V V be a linear endomorphism. We

More information

SUPPLEMENT TO CHAPTERS VII/VIII

SUPPLEMENT TO CHAPTERS VII/VIII SUPPLEMENT TO CHAPTERS VII/VIII The characteristic polynomial of an operator Let A M n,n (F ) be an n n-matrix Then the characteristic polynomial of A is defined by: C A (x) = det(xi A) where I denotes

More information

LINEAR ALGEBRA MICHAEL PENKAVA

LINEAR ALGEBRA MICHAEL PENKAVA LINEAR ALGEBRA MICHAEL PENKAVA 1. Linear Maps Definition 1.1. If V and W are vector spaces over the same field K, then a map λ : V W is called a linear map if it satisfies the two conditions below: (1)

More information

2.3. VECTOR SPACES 25

2.3. VECTOR SPACES 25 2.3. VECTOR SPACES 25 2.3 Vector Spaces MATH 294 FALL 982 PRELIM # 3a 2.3. Let C[, ] denote the space of continuous functions defined on the interval [,] (i.e. f(x) is a member of C[, ] if f(x) is continuous

More information

Math 350 Fall 2011 Notes about inner product spaces. In this notes we state and prove some important properties of inner product spaces.

Math 350 Fall 2011 Notes about inner product spaces. In this notes we state and prove some important properties of inner product spaces. Math 350 Fall 2011 Notes about inner product spaces In this notes we state and prove some important properties of inner product spaces. First, recall the dot product on R n : if x, y R n, say x = (x 1,...,

More information

Linear Algebra. Rekha Santhanam. April 3, Johns Hopkins Univ. Rekha Santhanam (Johns Hopkins Univ.) Linear Algebra April 3, / 7

Linear Algebra. Rekha Santhanam. April 3, Johns Hopkins Univ. Rekha Santhanam (Johns Hopkins Univ.) Linear Algebra April 3, / 7 Linear Algebra Rekha Santhanam Johns Hopkins Univ. April 3, 2009 Rekha Santhanam (Johns Hopkins Univ.) Linear Algebra April 3, 2009 1 / 7 Dynamical Systems Denote owl and wood rat populations at time k

More information

= W z1 + W z2 and W z1 z 2

= W z1 + W z2 and W z1 z 2 Math 44 Fall 06 homework page Math 44 Fall 06 Darij Grinberg: homework set 8 due: Wed, 4 Dec 06 [Thanks to Hannah Brand for parts of the solutions] Exercise Recall that we defined the multiplication of

More information

Lecture notes - Math 110 Lec 002, Summer The reference [LADR] stands for Axler s Linear Algebra Done Right, 3rd edition.

Lecture notes - Math 110 Lec 002, Summer The reference [LADR] stands for Axler s Linear Algebra Done Right, 3rd edition. Lecture notes - Math 110 Lec 002, Summer 2016 BW The reference [LADR] stands for Axler s Linear Algebra Done Right, 3rd edition. 1 Contents 1 Sets and fields - 6/20 5 1.1 Set notation.................................

More information

Linear Algebra Highlights

Linear Algebra Highlights Linear Algebra Highlights Chapter 1 A linear equation in n variables is of the form a 1 x 1 + a 2 x 2 + + a n x n. We can have m equations in n variables, a system of linear equations, which we want to

More information

THE JORDAN-FORM PROOF MADE EASY

THE JORDAN-FORM PROOF MADE EASY THE JORDAN-FORM PROOF MADE EASY LEO LIVSHITS, GORDON MACDONALD, BEN MATHES, AND HEYDAR RADJAVI Abstract A derivation of the Jordan Canonical Form for linear transformations acting on finite dimensional

More information

Solutions to the August 2008 Qualifying Examination

Solutions to the August 2008 Qualifying Examination Solutions to the August 2008 Qualifying Examination Any student with questions regarding the solutions is encouraged to contact the Chair of the Qualifying Examination Committee. Arrangements will then

More information

Lecture Summaries for Linear Algebra M51A

Lecture Summaries for Linear Algebra M51A These lecture summaries may also be viewed online by clicking the L icon at the top right of any lecture screen. Lecture Summaries for Linear Algebra M51A refers to the section in the textbook. Lecture

More information

(a) II and III (b) I (c) I and III (d) I and II and III (e) None are true.

(a) II and III (b) I (c) I and III (d) I and II and III (e) None are true. 1 Which of the following statements is always true? I The null space of an m n matrix is a subspace of R m II If the set B = {v 1,, v n } spans a vector space V and dimv = n, then B is a basis for V III

More information

Math 113 Winter 2013 Prof. Church Midterm Solutions

Math 113 Winter 2013 Prof. Church Midterm Solutions Math 113 Winter 2013 Prof. Church Midterm Solutions Name: Student ID: Signature: Question 1 (20 points). Let V be a finite-dimensional vector space, and let T L(V, W ). Assume that v 1,..., v n is a basis

More information

MATH 221: SOLUTIONS TO SELECTED HOMEWORK PROBLEMS

MATH 221: SOLUTIONS TO SELECTED HOMEWORK PROBLEMS MATH 221: SOLUTIONS TO SELECTED HOMEWORK PROBLEMS 1. HW 1: Due September 4 1.1.21. Suppose v, w R n and c is a scalar. Prove that Span(v + cw, w) = Span(v, w). We must prove two things: that every element

More information