Linear Algebra II. Course No Spring 2007 Michael Stoll

Transcription

1 Linear Algebra II Course No Spring 2007 Michael Stoll With some additions by Ronald van Luijk, 2016

2

3 Contents 1. Review of Eigenvalues, Eigenvectors and Characteristic Polynomial 2 2. Direct Sums of Subspaces 2 3. The Cayley-Hamilton Theorem and the Minimal Polynomial 7 4. The Structure of Nilpotent Endomorphisms The Jordan Normal Form Theorem The Dual Vector Space Norms on Real Vector Spaces Bilinear Forms Inner Product Spaces Orthogonal Diagonalization External Direct Sums The Tensor Product Symmetric and Alternating Products 61 References 67 Index of notation 68

4 2 1. Review of Eigenvalues, Eigenvectors and Characteristic Polynomial Recall the topics we finished Linear Algebra I with. We were discussing eigenvalues and eigenvectors of endomorphisms and square matrices, and the question when they are diagonalizable. Foryourconvenience,Iwillrepeatherethemostrelevant definitions and results. Let V be a finite-dimensional F -vector space, dim V = n, andletf : V! V be an endomorphism. Then for 2 F,the -eigenspace of f was defined to be E (f) ={v 2 V : f(v) = v} =ker(f id V ). is an eigenvalue of f if E (f) 6= {0}, i.e.,ifthereis06= v 2 V such that f(v) = v. Suchavectorv is called an eigenvector of f for the eigenvalue. The eigenvalues are exactly the roots (in F )ofthecharacteristic polynomial of f, P f (x) =det(x id V f), which is a monic polynomial of degree n with coe cients in F. The geometric multiplicity of as an eigenvalue of f is defined to be the dimension of the -eigenspace, whereas the algebraic multiplicity of as an eigenvalue of f is defined to be its multiplicity as a root of the characteristic polynomial. The endomorphism f is said to be diagonalizable if there exists a basis of V consisting of eigenvectors of f. Thematrixrepresentingf relative to this basis is then a diagonal matrix, with the various eigenvalues appearing on the diagonal. Since n n matrices can be identified with endomorphisms F n! F n,allnotions and results makes sense for square matrices, too. A matrix A 2 Mat(n, F )is diagonalizable if and only if it is similar to a diagonal matrix, i.e., if there is an invertible matrix P 2 Mat(n, F )suchthatp 1 AP is diagonal. It is an important fact that the geometric multiplicity of an eigenvalue cannot exceed its algebraic multiplicity. An endomorphism or square matrix is diagonalizable if and only if the sum of the geometric multiplicities of all eigenvalues equals the dimension of the space. This in turn is equivalent to the two conditions (a) the characteristic polynomial is a product of linear factors, and (b) for each eigenvalue, algebraic and geometric multiplicities agree. For example, both conditions are satisfied if P f is the product of n distinct monic linear factors. 2. Direct Sums of Subspaces The proof of the Jordan Normal Form Theorem, which is one of our goals, uses the idea to split the vector space V into subspaces on which the endomorphism can be more easily described. In order to make this precise, we introduce the notion of direct sum of linear subspaces of V Definition. Suppose I is an index set and U i V (for i 2 I) arelinear subspaces of a vector space V satisfying 0 1 (1) U j X A = {0} i2i\{j} for all j 2 I. Then we write L i2i U i for the subspace P i2i U i of V,andwecall this sum the direct sum of the subspaces U i. Whenever we use this notation, the U i

5 hypothesis (1) is implied. If I = {1, 2,...,n}, thenwealsowriteu 1 U 2 U n Lemma. Let V be a vector space, and U i V (for i 2 I) linear subspaces. Then the following statements are equivalent. (1) Every v 2 V can be written uniquely as v = P i2i u i with u i 2 U i for all i 2 I (and only finitely many u i 6=0). (2) P i2i U i = V, and for all j 2 I, we have U j \ P i2i\{j} U i = {0}. (3) If we have any basis B i of U i for each i 2 I, then these bases B i are pairwise disjoint, and the union S i2i B i forms a basis of V. (4) There exists a basis B i of U i for each i 2 I such that these bases B i are pairwise disjoint, and the union S i2i B i forms a basis of V. By statement (2) of this lemma, if these conditions are satisfied, then V is the direct sum of the subspaces U i,thatis,wehavev = L i2i U i. Proof. (1) ) (2) : Since every v 2 V can be written as a sum of elements of the U i,wehavev = P i2i U i. Now assume that v 2 U j \ P i6=j U i.thisgivestwo representations of v as v = u j = P i6=j u i.sincethereisonlyonewayofwritingv as a sum of u i s, this is only possible when v =0. (2) ) (3) : Since the elements of any basis are nonzero, and B i is contained in U i for all i, itfollowsfromu j \ P i2i\{j} U i = {0} that B i \ B j = ; for all i 6= j. Let B = S i2i B i. Since B i generates U i and P i U i = V,wefindthatB generates V. To show that B is linearly independent, consider a linear combination X X i2i For any fixed j 2 I, wecanwritethisas b2b i i,bb =0. U j 3 u j = X b2b j j,bb = X i6=j X b2b i i,bb 2 X U i. By (2), this implies that u j =0. SinceB j is a basis of U j,thisisonlypossiblewhen j,b =0forallb 2 B j. Since j 2 I was arbitrary, this shows that all coe cients vanish. (3) ) (4) : This follows by choosing any basis B i for U i (see Remark 2.3). (4) ) (1) : Take a basis B i for U i for each i 2 I. Write v 2 V as a linear combination of the basis elements in S i B i.sinceb i is a basis of U i,wemaywrite the part of the linear combination coming from B i as u i,whichyieldsv = P i u i with u i 2 U i. To see that the u i are unique, we note that the u i can be written as linear combinations of elements in B i ; the sum v = P i u i is then a linear combination of elements in S i B i,whichhastobethesameastheoriginallinear combination, because S i B i is a basis for V. It follows that indeed all the u i are uniquely determined Remark. The proof of the implication (3) ) (4) implicitly assumes the existence of a basis B i for each U i. The existence of a basis B i for U i is clear when U i is finite-dimensional, but for infinite-dimensional vector spaces this is more subtle. Using Zorn s Lemma, which is equivalent to the Axiom of Choice of Set Theory, one can prove that all vector spaces do indeed have a basis. See Appendix D of Linear Algebra I, 2015 edition (or later). We will use this more often. i6=j

6 Remark. If U 1 and U 2 are linear subspaces of the vector space V, thenstatement V = U 1 U 2 is equivalent to U 1 and U 2 being complementary subspaces Lemma. Suppose V is a vector space with subspaces U and U 0 such that V = U U 0. If U 1,...,U r are subspaces of U with U = U 1 U r and U 0 1,...,U 0 s are subspaces of U 0 with U 0 = U 0 1 U 0 s, then we have V = U 1 U r U 0 1 U 0 s. Proof. This follows most easily from part (1) of Lemma 2.2. The converse of this lemma is trivial in the sense that if we have V = U 1 U r U 0 1 U 0 s, then apparently the r + s subspaces U 1,...,U r,u 0 1,...,U 0 s satisfy the hypothesis (1), which implies that also the r subspaces U 1,...,U r satisfy this hypothesis, as well as the subspaces U 0 1,...,U 0 s;thenalsothetwosubspacesu = U 1 U r and U 0 = U U 0 s together satisfy the hypothesis and we have V = U U 0. In other words, we may write (U 1 U r ) (U 0 1 U 0 s)=u 1 U r U 0 1 U 0 s in the sense that if all the implied conditions of the form (1) are satisfied for one side of the equality, then the same holds for the other side, and the (direct) sums are then equal. In particular, we have U 1 (U 2 U r )=U 1 U r. The following lemma states that if two subspaces intersect each other trivially, then one can be extended to a complementary space of the other. Its proof also suggests how we can do the extension explicitly Lemma. Let U and U 0 be subspaces of a finite-dimensional vector space V satisfying U \ U 0 = {0}. Then there exists a subspace W V with U 0 W that is a complementary subspace of U in V. Proof. Let (u 1,...,u r )beabasisforu and (v 1,...,v s )abasisforu 0. Then Lemma 2.2 we have a basis (u 1,...,u r,v 1,...,v s )foru + U 0 = U U 0. By the Basis Extension Theorem of Linear Algebra 1, we may extend this to a basis (u 1,...,u r,v 1,...,v s,w 1,...,w t )forv.wenowletw be the subspace generated by v 1,...,v s,w 1,...,w t.then(v 1,...,v s,w 1,...,w t )isabasisforw and clearly W contains U 0. By Lemma 2.2 we conclude that U and W are complementary spaces. Next, we discuss the relation between endomorphisms of V and endomorphisms between the U i Lemma and Definition. Let V be a vector space with linear subspaces U i (i 2 I) such that V = L i2i U i. For each i 2 I, let f i : U i! U i be an endomorphism. Then there is a unique endomorphism f : V! V such that f Ui = f i for all i 2 I. We call f the direct sum of the f i and write f = L i2i f i.

7 Proof. Let v 2 V. Then we have v = P i u i as above, therefore the only way to define f is by f(v) = P i f i(u i ). This proves uniqueness. Since the u i in the representation of v above are unique, f is a well-defined map, and it is clear that f is linear, so f is an endomorphism of V Remark. If in the situation of Definition 2.7, V is finite-dimensional and we choose a basis B of V that is the concatenation of bases B i of the U i,thenthe matrix representing f relative to B will be a block diagonal matrix, where the diagonal blocks are the matrices representing the f i relative to the bases B i of the U i. In this finite-dimensional case the number of indices i 2 I for which U i is nonzero is finite, and it follows that the characteristic polynomial P f equals P f = Y i2i In particular, we have det f = Q i2i det f i,andtrf = P i2i Tr f i for the determinant and the trace Remark. An endomorphism f : V! V is diagonalisable if and only if V is the direct sum of the eigenspaces of f Lemma. Let V be a vector space with linear subspaces U i (i 2 I) such that V = L i2i U i. Let f : V! V be an endomorphism. Then there are endomorphims f i : U i! U i for i 2 I such that f = L i2i f i if and only if each U i is invariant under f (or f-invariant), i.e., f(u i ) U i. Proof. If f = L i f i, then f i = f Ui, hence f(u i ) = f Ui (U i ) = f i (U i ) U i. Conversely, suppose that f(u i ) U i. Then we can define f i : U i! U i to be the restriction of f to U i ;itisthenclearthatf i is an endomorphism of U i and that f equals L i f i,asthetwocoincideonallthesubspacesu i,whichtogether generate V. We now come to a relation between splittings of f as a direct sum and the characteristic or minimal polynomial of f. We call two polynomials p 1 (x) andp 2 (x) coprime if there are polynomials a 1 (x) and a 2 (x) such that a 1 (x)p 1 (x)+a 2 (x)p 2 (x) = Lemma. Let V be a vector space and f : V! V an endomorphism. Let p(x) =p 1 (x)p 2 (x) be a polynomial such that p(f) =0and such that p 1 (x) and p 2 (x) are coprime. Let U i =ker p i (f), for i =1, 2. Then V = U 1 U 2 and the U i are f-invariant. In particular, f = f 1 f 2, where f i = f Ui. Moreover, we have U 1 =im p 2 (f) and U 2 =im p 1 (f). Proof. Set K 1 =im p 2 (f) and K 2 =im p 1 (f).wefirstshowthatk i U i for i =1, 2. Let v 2 K 1 =im p 2 (f),sov = p 2 (f) (u) forsomeu 2 V.Then p 1 (f) (v) = p 1 (f) p 2 (f) (u) = p 1 (f)p 2 (f) (u) = p(f) (u) =0, so K 1 = im p 2 (f) ker p 1 (f) = U 1. The statement for i = 2 follows by symmetry. Now we show that U 1 \ U 2 = {0}. So let v 2 U 1 \ U 2. Then p 1 (f) (v) = p 2 (f) (v) = 0. Using id V =1(f) = a 1 (x)p 1 (x)+a 2 (x)p 2 (x) (f) =a 1 (f) p 1 (f)+a 2 (f) p 2 (f), P fi. 5

8 6 we see that v = a 1 (f) p 1 (f) (v) + a 2 (f) p 2 (f) (v) = a 1 (f) (0) + a 2 (f) (0) = 0. Next, we show that K 1 + K 2 = V. Using the same relation above, and the fact that p i (f) anda i (f) commute,wefindforv 2 V arbitrary that v = p 1 (f) a 1 (f) (v) + p 2 (f) a 2 (f) (v) 2 im p 1 (f) +im p 2 (f). These statements together imply that K i = U i for i =1, 2, and V = U 1 U 2. Indeed, let v 2 U 1. We can write v = v 1 + v 2 with v i 2 K i. Then U 1 3 v v 1 = v 2 2 U 2,butU 1 \ U 2 = {0}, sov = v 1 2 K 1. Finally, we have to show that U 1 and U 2 are f-invariant. So let (e.g.) v 2 U 1. Since f commutes with p 1 (f), we have p 1 (f) f(v) = p 1 (f) f (v) = f p 1 (f) (v) =f p 1 (f) (v) = f(0) = 0, (since v 2 U 1 =ker p 1 (f) ), hence f(v) 2 U 1 as well Proposition. Let V be a vector space and f : V! V an endomorphism. Let p(x) =p 1 (x)p 2 (x) p k (x) be a polynomial such that p(f) =0and such that the factors p i (x) are coprime in pairs. Let U i =ker p i (f). Then V = U 1 U k and the U i are f-invariant. In particular, f = f 1 f k, where f i = f Ui. Proof. We proceed by induction on k. Thecasek =1istrivial. Soletk 2, and denote q(x) =p 2 (x) p k (x). Then I claim that p 1 (x) andq(x) arecoprime. To see this, note that by assumption, we can write, for i =2,...,k, Multiplying these equations, we obtain a i (x)p 1 (x)+b i (x)p i (x) =1. A(x)p 1 (x)+b 2 (x) b k (x)q(x) =1; note that all the terms except b 2 (x) b k (x)q(x) thatwegetwhenexpandingthe product of the left hand sides contains a factor p 1 (x). We can then apply Lemma 2.11 to p(x) =p 1 (x)q(x) andfindthatv = U 1 U 0 and f = f 1 f 0 with U 1 =ker p 1 (f), f 1 = f U1,andU 0 =ker q(f), f 0 = f U 0. In particular, q(f 0 )=0. Byinduction,wethenknowthatU 0 = U 2 U k with U j =ker p j (f 0 ) and f 0 = f 2 f k,wheref j = f 0 Uj,forj =2,...,k. Finally, ker p j (f 0 ) =ker p j (f) (since the latter is contained in U 0 )andf j = f 0 Uj = f Uj, so that we obtain the desired conclusion from Lemma 2.5. The following little lemma about polynomials is convenient if we want to apply Lemma Lemma. If p(x) is a polynomial (over F ) and 2 F such that p( ) 6= 0, then (x ) m and p(x) are coprime for all m 1. Proof. First, consider m = 1. Let q(x) = p(x) p( ) 1;

9 7 this is a polynomial such that q( )=0. Therefore,wecanwriteq(x) =(x with some polynomial r(x). This gives us r(x)(x )+ 1 p(x) =1. p( ) Now, taking the mth power on both sides, we obtain an equation r(x) m (x ) m + a(x)p(x) =1. )r(x) 3. The Cayley-Hamilton Theorem and the Minimal Polynomial Let A 2 Mat(n, F ). We know that Mat(n, F )isanf-vector space of dimension n 2. Therefore, the elements I, A, A 2,...,A n2 cannot be linearly independent (because their number exceeds the dimension). If we define p(a) intheobviouswayforp apolynomialwithcoe cientsinf (as we already did in the previous chapter), then we can deduce that there is a (non-zero) polynomial p of degree at most n 2 such that p(a) = 0 (0 here is the zero matrix). In fact, much more is true. Consider a diagonal matrix D =diag( 1, 2,..., n). (This notation is supposed to mean that j is the (j, j) entryofd; the o -diagonal entries are zero, of course.) Its characteristic polynomial is P D (x) =(x 1)(x 2) (x n). Since the diagonal entries are roots of P D,wealsohaveP D (D) =0.Moregenerally, consider a diagonalizable matrix A. Then there is an invertible matrix Q such that D = Q 1 AQ is diagonal. Since (Exercise!) p(q 1 AQ) =Q 1 p(a)q for p a polynomial, we find 0=P D (D) =Q 1 P D (A)Q = Q 1 P A (A)Q =) P A (A) =0. (Recall that P A = P D similarmatriceshavethesamecharacteristicpolynomial.) The following theorem states that this is true for all square matrices (or endomorphisms of finite-dimensional vector spaces) Theorem (Cayley-Hamilton). Let A 2 Mat(n, F ). Then P A (A) =0. Proof. Here is a simple, but wrong proof. By definition, P A (x) =det(xi A), so, plugging in A for x, wehavep A (A) = det(ai A) = det(a A) =det(0)=0. (Exercise: find the mistake!) For the correct proof, we need to consider matrices whose entries are polynomials. Since polynomials satisfy the field axioms except for the existence of inverses, we can perform all operations that do not require divisions. This includes addition, multiplication and determinants; in particular, we can use the adjugate matrix. Let B = xi A, thendet(b) =P A (x). Let B be the adjugate matrix; then we still have BB =det(b)i. Theentriesof B come from determinants of (n 1) (n 1) submatrices of B, therefore they are polynomials of degree at most n 1. We can then write B = x n 1 B n 1 + x n 2 B n xb 1 + B 0, and we have the equality (of matrices with polynomial entries) (x n 1 B n 1 +x n 2 B n 2 + +B 0 )(xi A) =P A (x)i =(x n +b n 1 x n 1 + +b 0 )I,

10 8 where we have set P A (x) =x n + b n 1 x n b 0.Expandingthelefthandside and comparing coe cients of like powers of x, wefindtherelations B n 1 = I, B n 2 B n 1 A = b n 1 I,..., B 0 B 1 A = b 1 I, B 0 A = b 0 I. We multiply these from the right by A n, A n 3.2. Remarks. B n 1 A n = A n B n 2 A n 1 B n 1 A n = b n 1 A n 1... B 0 A B 1 A 2 = b 1 A B 0 A = b 0 I 0 = P A (A) 1,...,A, I, respectively,andadd: (1) The reason why we cannot simply plug in A for x in the identity B (xi A) =P A (x)i is that whereas x (as a scalar) commutes with the matrices occurring as coe cients of powers of x, itisnotaprioriclearthata does so, too. We will discuss this in more detail in the Introductory Algebra course, where polynomial rings will be studied in some detail. (2) Another idea of proof (and maybe easier to grasp) is to say that a generic matrix is diagonalizable (if we assume F to be algebraically closed... ), hence the statement holds for most matrices. Since it is just a bunch of polynomial relations between the matrix entries, it then must hold for all matrices. This can indeed be turned into a proof, but unfortunately, this requires rather advanced tools from algebra. (3) Of course, the statement of the theorem remains true for endomorphisms. Let f : V! V be an endomorphism of the finite-dimensional F -vector space V,thenP f (f) =0(whichisthezeroendomorphisminthiscase). For evaluating the polynomial at f, wehavetointerpretf n as the n-fold composition f f f, andf 0 =id V. Our next goal is to define the minimal polynomial of a matrix or endomorphism, as the monic polynomial of smallest degree that has the matrix or endomorphism as a root. However, we need to know a few more facts about polynomials in order to see that this definition makes sense Lemma (Polynomial Division). Let f and g be polynomials, with g monic. Then there are unique polynomials q and r such that r =0or deg(r) < deg(g) and such that f = qg + r. Proof. We first prove existence, by induction on the degree of f. If deg(f) < deg(g), then we take q =0andr = f. So we now assume that m =deg(f) deg(g) =n, f = a m x m + + a 0.Letf 0 = f a m x m n g,then(sinceg = x n +...) deg(f 0 ) < deg(f). By the induction hypothesis, there are q 0 and r such that deg(r) < deg(g) orr =0andsuchthatf 0 = q 0 g +r. Thenf =(q 0 +a m x m n )g +r. (This proof leads to the well-known algorithm for polynomial long division.)

11 As to uniqueness, suppose we have f = qg + r = q 0 g + r 0,withr and r 0 both of degree less than deg(g) orzero.then (q q 0 )g = r 0 r. If q 6= q 0,thenthedegreeofthelefthandsideisatleastdeg(g), but the degree of the right hand side is smaller, hence this is not possible. So q = q 0,andtherefore r = r 0,too. Taking g = x, thisprovidesadi erentproofforcasek =1ofExample8.4of Linear Algebra I, 2015 edition (or later) Lemma and Definition. Let A 2 Mat(n, F ). There is a unique monic polynomial M A of minimal degree such that M A (A) =0. If p is any polynomial satisfying p(a) =0, then p is divisible by M A (as a polynomial). This polynomial M A is called the minimal (or minimum) polynomial of A. Similarly, we define the minimal polynomial M f of an endomorphism f of a finitedimensional vector space. Proof. It is clear that monic polynomials p with p(a) =0exist(bytheCayley- Hamilton Theorem 3.1, we can take p = P A ). So there will be one such polynomial of minimal degree. Now assume p and p 0 were two such monic polynomials of (the same) minimal degree with p(a) =p 0 (A) =0. Thenwewouldhave(p p 0 )(A) = p(a) p 0 (A) =0. Ifp 6= p 0,thenwecandividep p 0 by its leading coe cient, leading to a monic polynomial q of smaller degree than p and p 0 with q(a) =0, contradicting the minimality of the degree. Now let p be any polynomial such that p(a) = 0. By Lemma 3.3, there are polynomials q and r, deg(r) < deg(m A )orr = 0, such that p = qm A + r. Plugging in A, wefindthat 0=p(A) =q(a)m A (A)+r(A) =q(a) 0+r(A) =r(a). If r 6= 0,thendeg(r) < deg(m A ), but the degree of M A is the minimal possible degree for a polynomial that vanishes on A, sowehaveacontradiction. Therefore r =0andhencep = qm A Remark. In Introductory Algebra, you will learn that the set of polynomials as discussed in the lemma forms an ideal and that the polynomial ring is a principal ideal domain, which means that every ideal consists of the multiples of some fixed polynomial. The proof is exactly the same as for the lemma. By Lemma 3.4, the minimal polynomial divides the characteristic polynomial. As a simple example, consider the identity matrix I n.itscharacteristicpolynomialis (x 1) n,whereasitsminimalpolynomialisx 1. In some sense, this is typical, as the following result shows Proposition. Let A 2 Mat(n, F ) and 2 F. If is a root of the characteristic polynomial of A, then it is also a root of the minimal polynomial of A. In other words, both polynomials have the same linear factors. Proof. If P A ( )=0,then is an eigenvalue of A, sothereis06= v 2 F n such that Av = v. SettingM A (x) =a m x m + + a 0,wefind mx mx 0=M A (A)v = a j A j v = j v = M A ( )v. j=0 j=0 a j 9

12 10 (Note that the terms in this chain of equalities are vectors.) Since v 6= 0,this implies M A ( )=0. By Lemma 3.4, we know that each root of M A is a root of P A,andwehavejust shown the converse. So both polynomials have the same linear factors Remark. If F is algebraically closed (i.e., every non-zero polynomial is a product of linear factors), this shows that P A is a multiple of M A,andM k A is a multiple of P A when k is large enough. In fact, the latter statement is true for general fields F (and can be interpreted as saying that both polynomials have the same irreducible factors). For the proof, one replaces F by a larger field F 0 such that both polynomials split into linear factors over F 0. That this can always be done is shown in Introductory Algebra. One nice property of the minimal polynomial is that it provides another criterion for diagonalizability Proposition. Let A 2 Mat(n, F ). Then A is diagonalizable if and only if its minimal polynomial M A is a product of distinct monic linear factors. Proof. First assume that A is diagonalizable. It is easy to see that similar matrices have the same minimal polynomial (Exercise), so we can as well assume that A is already diagonal. But for a diagonal matrix, the minimal polynomial is just the product of factors x, where runs through the distinct diagonal entries. (It is the monic polynomial of smallest degree that has all diagonal entries as roots.) Conversely, assume that M A (x) = (x 1) (x m) with 1,..., m 2 F distinct. The polynomials q i = x i (with 1 apple i apple m) arepairwisecoprime,so by Proposition 2.12 the eigenspaces U i = E i (A) =ker(a ii) =kerq i (A) satisfy F n = U 1 U m. This implies n = P m i=1 dim E (A), which in turn i (by Cor of Linear Algebra I, 2015 edition) impliesthata is diagonalizable Example. Consider the matrix 0 A A Is it diagonalizable? Its characteristic polynomial is clearly P A (x) =(x 1) 3,soitsminimalpolynomial must be (x 1) m for some m apple 3. Since A I 6= 0,m>1(infact,m =3),hence A is not diagonalizable. On the other hand, the matrix (for F = R, say) 0 B A has M B (x) =P B (x) =(x 1)(x 4)(x 6); B therefore is diagonalizable. Exercise: what happens for fields F of small characteristic?

13 11 4. The Structure of Nilpotent Endomorphisms 4.1. Definition. AmatrixA2Mat(n, F )issaidtobenilpotent, if A m =0for some m 1. Similarly, if V is a finite-dimensional vector space and f : V! V is an endomorphism, then f is said to be nilpotent if f m = f f f =0for {z } some m 1. m times It follows that the minimal polynomial of A or f is of the form x m,wherem is the smallest number that has the property required in the definition Proposition. A nilpotent matrix or endomorphism is diagonalizable if and only if it is zero. Proof. The minimal polynomial is x m. Prop. 3.6 then implies that the matrix or endomorphism is diagonalizable if and only if m = 1. But then the minimal polynomial is x, whichmeansthatthematrixorendomorphismiszero. The following result tells us more about the structure of nilpotent endomorphisms. It is the main ingredient to proving the existence of the Jordan Normal Form Theorem. Let V be an F -vector space, dim V = n, and let f : V! V be a nilpotent endomorphism. Then V has a basis (v 1,v 2,...,v n ) such that f(v j ) is either zero or v j+1. Proof. Let m be an integer such that f m = 0. Note that we have a chain of inclusions {0} =kerf 0 ker f 1 ker f 2 ker f m 1 ker f m = V. We prove by descending induction that for all j 2{0, 1,...,m} there are elements w 1,...,w s 2 V and non-negative integers e 1,...,e s, such that the sequence (2) w 1,f(w 1 ),...,f e 1 (w 1 ),w 2,f(w 2 ),...,f e 2 (w 2 ),...,w s,f(w s ),...,f es (w s ) is a basis of a complementary space U j of ker f j inside V and the elements f e 1+1 (w 1 ),...,f es+1 (w s )arecontainedinkerf j. For j = m we have ker f m = V,sowemaytakes =0andU m = {0}. Suppose 0 apple j < m and suppose that we have elements w 1,...,w t 2 V and integers e 1,...,e s,suchthatthesequenceanalogousto(2)isabasisforacomplementary subspace U j+1 of ker f j+1 inside V,andtheelementsf e 1+1 (w 1 ),...,f es+1 (w s )are contained in ker f j+1. We claim that if any scalars 1,..., s satisfy P s i=1 if ei+1 (w i ) 2 ker f j,thenwe have 1 = = s =0. Indeed,setz = P s i=1 if e i (w i ). Then the assumption states that f(z) 2 ker f j,sowehavez 2 ker f j+1. But z is also a linear combination of elements in a basis for U j+1,sowehavez 2 U j+1 \ ker f j+1 = {0} and thus z =0,whichinturnimplies 1 = = s =0. The claim implies in particular that the elements f ei+1 (w i )fori =1,...,s are linearly independent, so they form a basis for the subspace W j ker f j+1 that they generate. The claim also implies W j \ ker f j = {0}, sow j can be extended to a complementary space X j of ker f j inside ker f j+1,andthebasisforw j can be extended to a basis (f e1+1 (w 1 ),...,f es+1 (w s ),w s+1,...,w t )forx j. Set U j = U j+1 X j.thenwehave V = U j+1 ker f j+1 = U j+1 (X j ker f j )=(U j+1 X j ) ker f j = U j ker f j,

14 12 so U j is a complementary space of ker f j in V. The union of the two bases for U j+1 and X j gives a basis for U j,whichwecanrearrangeto w 1,f(w 1 ),...,f e 1+1 (w 1 ),w 2,f(w 2 ),...,f e 2+1 (w 2 ),...,w s,f(w s ),...,f es+1 (w s ),w s+1,...,w t. This finishes the induction step, as it is of the same form as (2), with t substituted for s, withe i +1substitutedfore i (for 1 apple i apple s), and with 0 substituted for e i (for s<i<apple t). The statement of the Theorem now follows from the case j = 0, as the only complementary subspace of ker f 0 = {0} is V,andwemaydenotetheelementsof (2) for j =0byv 1,v 2,...,v n Remark. The matrix A =(a ij )representingf with respect to v n,...,v 2,v 1, where v 1,...,v n is a basis as in Thm. 4.3 above, has all entries zero except a j,j+1 = 1iff(v n j )=v n+1 j.thereforeais a block diagonal matrix 0 1 B B A = 2 0 C. A 0 0 B n k with blocks of the form B j =. B A Note that we reversed the order of the basis elements! 4.5. Corollary. A nilpotent matrix is similar to a matrix of the form just described. Proof. This is clear from our discussion Corollary. A matrix A 2 Mat(n, F ) is nilpotent if and only if P A (x) =x n. Proof. If P A (x) =x n,thena n = 0 by the Cayley-Hamilton Theorem 3.1, hence A is nilpotent. Conversely, if A is nilpotent, then it is similar to a matrix of the form above, which visibly has characteristic polynomial x n Remark. The statement of Cor. 4.6 would also follow from the fact that P A (x) dividessomepowerofm A (x) =x m, see Remark 3.7. However, we have proved this only in the case that P A (x) splitsintolinearfactors(whichweknow is true, but only after the fact).

15 4.8. Remark. The proof of Theorem 4.3 gives an e cient algorithm to construct such a special basis. (1) Compute an m with f m =0. (2) Start with j = m and U m = {0} and s =0andtheemptybasisforU m. (3) Decrease j by 1. We now have a basis w 1,f(w 1 ),...,f e 1 (w 1 ),w 2,f(w 2 ),...,f e 2 (w 2 ),...,w s,f(w s ),...,f es (w s ) for a complement U j+1 of ker(f j+1 )inv such that f e j+1 (w i ) 2 ker(f j+1 ) for 1 apple i apple s. (4) Extend f e 1+1 (w 1 ),...,f es+1 (w s )toabasis f e 1+1 (w 1 ),...,f es+1 (w s ),w s+1,...,w t of a complementary space of ker f j in ker f j+1 (see Lemma 2.6). (5a) For 1 apple i apple s, increasee i by 1. (5b) For s<iapple t, sete i =0. (5c) Set s equal to t. (6) If j =0,thenwearedone. Ifj>0, go back to (3). The last proof of Theorem 4.3 shows that step (4) is possible, that the condition after step (3) always holds, and that in the end we end up with a basis of the desired form Example. Consider 0 A A 2 Mat(3, R) We find 0 A A and A 3 =0,soAisnilpotent. Let us find a basis as given in Thm The first step in the process comes down to finding a complementary subspace of ker(a 2 )=L (2, 0, 1) >, ( 1, 1, 0) >. We can take (1, 0, 0) >,forexample,asthe basis of a complement. This will be w 1 in the notation of the proof above. We then have Aw 1 =(3, 1, 2) > and A 2 w 1 =( 1, 1, 1) >,andthesethreealready form a basis. Reversing the order, we get A A The Jordan Normal Form Theorem In this section, we will formulate and prove the Jordan Normal Form Theorem, which will tell us that any matrix whose characteristic polynomial is a product of linear factors is similar to a matrix of a very special near-diagonal form. Now we can feed this into Prop