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 of the notation for the iterated kernels and many of the ideas for the proof of Jordan s Decomposition Theorems are borrowed from [1. The flavor to functional analysis (Riesz theory) is taken from [2. Finally, notation has been adapted from [3 to fit in what I was teaching in MATH 672 in the Fall 213 semester. 1 Preliminary notions Definition 1. Let A M(n; F), and let V be a subspace of F n = F n c A invariant if A u V, u V.. We say that V is Examples. If λ is an eigenvalue of A, then ker(a λi) = {u F n : A u = λ u} is A invariant. More generally Proof. If u ker(a λi) j, then ker(a λi) j is A invariant j. (A λ I) j u = = = A (A λ I) j u ( ) = (A λ I) j A u To prove (*), we just have to notice that = Au ker(a λi) j. (A λ I) j = and therefore A (A λ I) j = (A λ I) j A. j ( ) j ( λ) j i A i, i i= Definition 2. If A M(n; F) and F n = V W, where V and W are A invariant, we say that F n = V W is an A invariant decomposition. The same definition can be extended to F n = V 1... V k, where V j is A invariant for all j. 1

Example. If A is diagonalizable and λ 1,..., λ k are the distinct eigenvalues of A, then F n = ker(a λ 1 I)... ker(a λ k I) is an A invariant decomposition. Also if Av j = µ j v j, j and {v 1,..., v n } is a basis for F n, then F n = span[v 1 span[v 2... span[v n is an A invariant decomposition. The associated matrix. If F n = V W is an A invariant decomposition, and we construct a basis for F n as follows: and we build the matrix P = v 1,..., v }{{} l, v l+1,..., v }{{ n } basis for V basis for W v 1... v l v l+1... v n, then [ P 1 Al l A P = A n l n l In general, invariant decomposition produce block diagonal matrices. Note that if A 1 P 1 A 2 A P =... A k where A j is a square matrix for all j, then: (a) the corresponding partition of the columns of P creates an A invariant decomposition of F n in k subspaces; (b) the characteristic polynomial of A is the product of the characteristic polynomials of the matrices A j. First goal. In a first step we will try to find a matrix P such that A 1 P 1 A 2 A P =... A k where χ Aj (x) = (λ j x) m j λ i λ j i j. 2

This will be possible for any matrix such that χ A (x) = (λ 1 x) m 1... (λ k x) m k m 1 +... + m k = n. We will have thus created an A invariant decomposition F n = V 1 V 2... V k, where dim V j = m j. From now on, we will restrict our attention to the case F = C. The case where A is a real matrix whose characteristic polynomial has only real roots will follow as a particular case. 2 Iterated kernels Notation. From now on we will consider a matrix whose characteristic polynomial is χ A (x) = (λ 1 x) m 1... (λ k x) m k, with pairwise different roots {λ 1,..., λ k } with multiplicities m j 1. Definition 3. Let A M(n; C) and λ be an eigenvalue of A. The subspaces E j (λ) = ker(a λ I) j, j 1, are called iterated kernels. We denote E (λ) = ker( I ) =. Elementary properties. (1) E 1 (λ) is the set of all eigenvectors associated to λ and the zero vector (which is never considered an eigenvector). (2) E j (λ) E j+1 (λ) for all j. Proof. It follows from the following straightforward argument (A λ I) j u = = (A λ I) j+1 u =. (3) If E l (λ) = E l+1 (λ), then E l+1 (λ) = E l+2 (λ). Consequently E l (λ) = E l+1 (λ) = E l+2 (λ) =.... Proof. Let u be such that (A λ I) l+2 u = and define v = (A λ I)u. Then (A λi) l+1 v = (A λi) l+2 u =, which implies that v E l+2 (λ) = E l+1 (λ), and therefore (A λi) l+1 u = (A λi) l v =. This argument shows then that E l+2 (λ) E l+1 (λ) E l+2 (λ). (4) The spaces E j (λ) are A invariant. 3

Not so elementary properties. Section 3. The proofs for the following results will be given in (5) For all j 1 dim E j+1 (λ) dim E j (λ) dim E j (λ) dim E j 1 (λ). (6) Let λ 1,..., λ k be eigenvalues of A. Let j 1,..., j k 1. Then, the following sum of subspaces is direct: E j1 (λ 1 ) E j2 (λ 2 )... E jk (λ k ). In other words, the spaces E j1 (λ 1 ),..., E jk (λ k ) are independent. (7) The maximum dimension of the iterated kernels E k (λ) is the multiplicity of λ as a root of the characteristic polynomial. In other words, for every eigenvalue λ, there exists l such that E l (λ) = E l+1 (λ), dim E l (λ) = m λ, where m λ is the algebraic multiplicity of λ as a root of χ A. On the dimensions of the iterated kernels. Let n j = dim E j (λ) = dim ker(a λ I) j, j (note that n = ). The preceding properties imply that this sequence of integers satisfies the following properties: (1) = n < n 1 <... < n l = n l+1 =..., (2) n l n l 1 n l 1 n l 2... n 2 n 1 n 1 n = n 1 (3) n l = m λ. The sequence of iterated kernels is then E (λ) E 1 (λ) E 2 (λ)... E l (λ) = E l+1 (λ) =... dim : = n < n 1 < n 2 <... < n l = n l+1 =... A simple but important observation. If B = P 1 AP, then u ker(b λi) j P u ker(a λi) j. Therefore, the dimensions of the iterated kernels are invariant by similarity transformations. This can be also explained in different words: the dimensions of the iterated kernels depend on the operator and not on the particular matrix representation of the operator. 4

Theorem 1 (First Jordan decomposition theorem). If χ A (x) = (λ 1 x) m 1 (λ 2 x) m 2... (λ k x) m k, then there exist integers l j such that C n = E l1 (λ 1 )... E lk (λ k ) and there is a change of basis such that P 1 A P = A 1..., where χ Ap (x) = (λ p x) mp Furthermore, the dimensions of the iterated kernels of the submatrices A p coincide with the dimensions of the kernels E j (λ p ) for the matrix A, that is, A k p. dim ker(a p λ p I) j = dim ker(a λ p I) j j. Proof. The first assertion is a direct consequence of properties (6) and (7). Taking bases of the iterated kernels E lj (λ j ) and placing them as columns of a matrix P we reach a block diagonal form. Let us show that the dimensions of the iterated kernels of A 1 coincide with the dimensions of E j (λ 1 ). Let B = P 1 A P. Since u ker(b λ 1 I) j P u E j (λ 1 ) E l1 (λ 1 ) = span[p 1, p 2,..., p m1 (p j are the column vectors of P ), then ker(b λ 1 I) j span[e 1, e 2,..., e m1 (e j are the canonical vectors). This means that elements of ker(b λ 1 I) j can only have its first m 1 entries non-vanishing. From that, it is obvious that [ v u ker(b λ 1 I) j u =, v ker(a 1 λ 1 I) j. The same proof applies to all other eigenvalues. Very relevant conclusion. Once the first Jordan Theorem has been proved, we can restrict our attention to matrices whose characterstic polynomial is of the form (λ 1 x) n, since this theorem allows us to separate the space into smaller subspaces where the operator/matrix only involves one eigenvalue. 3 Proofs 5

This section can be skipped in a first reading. Property (5). For all j 1 dim E j+1 (λ) dim E j (λ) dim E j (λ) dim E j 1 (λ). Proof. Let m = dim E j+1 (λ) dim E j (λ) and let us take independent vectors {v 1,..., v m } such that E j+1 (λ) = E j (λ) span[v 1,..., v m. This can be written in the following form: v 1,..., v m E j+1 (λ), ξ 1 v 1 +... + ξ m v m E j (λ) ξ 1 =... = ξ m =. Let now w i := (A λ I)v i. It is easy to verify that and therefore w 1,..., w m E j (λ), ξ 1 w 1 +... + ξ m w m E j 1 (λ) ξ 1 =... = ξ m =, dim span[w 1,..., w m = m span[w 1,..., w m E j (λ), span[w 1,..., w m E j 1 (λ) = {}. Therefore dim E j (λ) dim E j 1 (λ) m. (the vectors are independent), Lemma 1. If p(a)u = and q(a)u =, where p, q C[x, and we define r = g.c.d.(p, q), then r(a)u =. Proof. Both p and q have to be multiples of the minimal polynomial of u with respect to A. Therefore r is a multiple of this same polynomial which implies the result. (Recall that the minimal polynomial for a vector u with respect to a matrix A is the lowest order monic polynomial p such that p(a)u = and that if q(a)u =, then p divides q.) Property (6). Let λ 1,..., λ k be pairwise different eigenvalues of A. Let j 1,..., j k 1. The the following sum of spaces is direct: Proof. Let u 1,..., u k such that Then (A λ 1 I) j 1 u 1 =, E j1 (λ 1 ) E j2 (λ 2 )... E jk (λ k ). u 1 + u 2 +... + u k =, u i E ji (λ i ). (A λ 2 I) j 2... (A λ k I) j k u 1 = (A λ 2 I) j 2... (A λ k I) j k (u 2 +... + u k ) =. 6

Since ( ) g.c.d. (x λ 1 ) j 1, (x λ 2 ) j 2... (x λ k ) j k = 1, then u 1 =. Proceeding by induction we prove that u j = for all j. This proves that the subspaces are independent. Property (7a). For all j dim E j (λ) m λ, where χ A (x) = (x λ) m λ q(x) with q(λ). Proof. Let P be invertible such that its first m columns form a basis of E j (λ) (m = dim E j (λ)). Therefore [ P 1 Am C A P = m, A B m M(m; C). m If A m v = µ v, then P 1 A P [ v = µ [ v = A P [ v = µp [ v. Therefore w = P [ v E 1 (µ) E j (λ). If µ λ, then E 1 (µ) E j (λ) = {}, so w = and v =. This shows that χ Am (x) = (λ x) m. Since χ A (x) = χ Am (x)χ Bm (x) = (λ x) m χ Bm (x) = (λ x) m λ q(x), q(λ) it follows that m m λ. Property (7b). If E l+1 (λ) = E l (λ), then dim E l (λ) m λ. Proof. Let m = dim E l (λ). We take P as in the previous proof, so that [ P 1 Am C A P = m = B Ã, m where A m M(m; C). We are going to see that B m cannot have λ as an eigenvalue. Let us recall that for all j Therefore u ker(ã λ I)j P u ker(a λ I) j. ker(ã λ I)l = ker(ã λ I)l+k, ker(ã λ I)l = span[e 1,..., e m. k 7

The latter identity implies that the first m columns of (Ã λ I)l vanish. We can then write [ [ (Ã λ (Am λ I) I)l = l Ĉ m Ĉm (B m λ I) l = (B m λ I) l. Now, if (B m λi)v =, then (Ã λ I)2l [ v = [ Ĉm (B m λ I) l [ Ĉ m v (B m λ I) l v = [ Ĉm (B m λ I) l [ Ĉm v = which implies and therefore v =. (Ã λ I)l [ v = Nota. Properties (7a) and (7b) imply Property (7). However, Property (7b) is enough to show that dim E l (λ) = m(λ). The reason for this is that Property (6) implies n dim E l1 (λ 1 ) +... + dim E lk (λ k ) m 1 +... + m k = n which forces all the inequalities to be equalities. 4 Jordan blocks and Jordan matrices Definition 4. A Jordan block of order k associated to a value λ is a k k matrix of the form λ 1 λ 1 J k (λ) := λ....... 1 λ Properties. (1) The characteristic polynomial of J k (λ) is (λ x) k. Moreover, dim ker(j k (λ) λi) = 1. (2) The dimensions of the iterated kernels for J k (λ) are: { j, j = 1,..., k dim ker(j k (λ) λi) j = k, j k } = min(k, j). 8

Proof. Note that J k (λ) λi = 1 1...... 1 and take increasing powers of this matrix, noticing how the rank reduces by onw with each power. Otherwise, note that n 1 = 1 (the dimension of the space of eigenvectors is one) and by Property (5) of the iterated kernels the dimensions can only increase one by one until it reaches k. Definition 5. A Jordan matrix associated to a value λ is a block diagonal matrix whose blocks are Jordan blocks for the value λ: J k1 (λ) J k2 (λ)... J kr (λ) A general Jordan matrix is a block diagonal matrix, whose blocks are Jordan matrices associated to different values. Equivalently, it is a block diagonal matrix whose blocks are Jordan blocks. 5 Two combinatorial problems Objetivo. We will next show how to establish a one-to-one correspondence between the different possibilities for dimensions of iterated kernels for an m m Jordan matrix (with a single eigenvalue) and the possibilities of setting up the Jordan boxes until we fill the m m space. Problem (P). Find a sequence of integers satisfying (a) = n < n 1 <... < n l = n l+1 =..., (b) n l n l 1 n l 1 n l 2... n 2 n 1 n 1 n = n 1 (c) n l = m. Problem (Q). Find a sequence of non-negative integers q 1, q 2,..., q l,... such that (a) q l+1 = q l+2 =... = (b) q 1 + 2 q 2 + 3 q 3 +... + l q l = m 9

Remark. Before showing how Problem (P) and Problem (Q) are related, let us look ahead and explain what they are going to mean in the context of Jordan matrices. We also give the interpretation for the solutions of an intermediate problem: {n j } will be the dimensions of the iterated kernels; {p j } will be the number of Jordan blocks of size j j or higher; {q j } will be the number of j j Jordan blocks. The relation between both sequences will be done without taking into account what is the value of l where the solution of (P) stagnates and where the one of (Q) takes its last non-zero value. This value will be however the same in both sequences. From (P) to (Q). Given a solution to (P), we define p j = n j n j 1, j 1. This sequence satisfies =... = p l+1 < p l p l+1... p 1 = n 1. We next define q j = p j p j+1, j 1, so that q l = p l and q j = for all j > l. The sequence {q j } is a solution to (Q). From (Q) to (P). Given a solution to (Q), we define p j = q j + q j+1 +... + q l, j = 1,..., l, p j = j l, and then define a new sequence using a recurrence { n = n j = p j + n j 1, j 1. Then {n j } is a solution of (P). This shows that there is a bijection between solutions of (P) and (Q). Theorem 2. Let {n j } be a solution to (P) and let {q j } be the associated solution to (Q). Then the m m Jordan matrix associated to the value λ, built with q j blocks of size j j for all j satisfies dim E j (λ) = n j, j. 1

Proof. Let us consider the Jordan matrix constructed as explained in the statement. Since the matrix is block diagonal, it follows that dim E j (λ) = q l dim ker(j l (λ) λi) j + q l 1 dim ker(j l 1 (λ) λi) j +... + q 1 dim ker(j 1 (λ) λi) j = q l min(j, l) + q l 1 min(j, l 1) +... + q 1 min(j, 1) = j (q l + q l 1 +... + q j+1 ) + jq j + (j 1)q j 1 +... + 2q 2 + q 1 = j(p l + p l 1 p l +... + p j+1 p j+2 ) +j(p j p j+1 ) +... + 2(p 2 p 3 ) + (p 1 p 2 ) = j p j+1 j p j+1 + p j + p j 1 +... + p 1 = n j. This proves the result. Conclusions. (1) For any solution of problem (P), there exists an m m Jordan matrix with a single eigenvalue such that dim E j (λ) = n j, j. (2) Consequently, given possible configurations of the dimensions of the iterated kernels associated to all eigenvalues, there exists a matrix (a Jordan matrix actually) whose iterated kernels have the given dimensions. (3) Two essentially different Jordan matrices (that cannot be obtained by permutation in the order of the blocks) are not similar. Proof. The block-configuration (how many blocks of each size for each eigenvalue) determines univocally the dimensions of the iterated kernels (different block configurations imply different dimensions of the kernels). However, the dimensions of the iterated kernel do not vary under similarity transformations. 6 Existence of a Jordan canonical form An observation. If P 1 A P = and u 1,..., u k are the first columns of P, then Au 1 = λu 1 [ Jk (λ) Au 2 = λu 2 + u 1. so for all j 2 Au k = λu k + u k 1 u j 1 = (A λi)u j. 11

Lemma 2. If V is a subspace satisfying E j 1 (λ) V E j (λ) and W = (A λi)v = {(A λi) u : u V}, then E j 2 (λ) W E j 1 (λ), dim W = dim V. Proof. There are three things to check: (a) W E j 1 (λ), which is quite obvious, since u E j (λ) = (A λi)u E j 1 (λ); (b) W E j 2 (λ) =, since if v W (so v = (A λi)u, for some u V), then (A λi) j 2 v = = (A λi) j 1 u = = u V E j 1 (λ) = ; (c) if u V, then (A λi)u = = u E 1 (λ) V E j 1 (λ) V =, so linear independence is preserved by multiplication by (A λi). The proof of Theorem 3 is quite technical. It can be skipped in a first reading. Theorem 3 (Second Jordan decomposition theorem). Every m m with characteristic polynomial (x λ) m is similar to a Jordan matrix associated to the value λ. Moreover, the dimensions of the Jordan boxes are given by the coefficients {q j } associated to the dimensions {n j } of the iterated kernels. Proof. Let E l (λ) = C m be the first iterated kernel of maximum dimension. Let V l = V l tal que E l (λ) = E l 1 (λ) V l, so that dim V l = n l n l 1 = p l = q l. Using Lemma 2 above E l 1 (λ) = E l 2 (λ) (A λi)vl Vl 1, }{{} V l 1 with Likewise dim V l 1 = (n l 1 n l 2 ) dim V l = p l 1 p l = q l 1. E l 2 (λ) = E l 3 (λ) (A λi)v l 1 Vl 2 = }{{} V l 2 = E l 3 (λ) (A λi) 2 V l (A λi)v l 1 V l 2, 12

with dim V l 2 = q l 2. Proceeding succesively, we can build a partition of E l (λ) E l (λ) = Vl (A λi)vl... (A λi) l 2 Vl (A λi) l 1 Vl Vl 1... (A λi) l 3 Vl 1 (A λi) l 2 Vl 1.. V2 (A λi)v2 V1 so that, dim Vj = q j, j = 1,..., l. Note that we can decompose E l ()λ) in two different ways: E l (λ) = V l V l 1... V 2 V 1 = W l W l 1... W 2 W 1, where V j = V j (A λi)v j+1... (A λi) l j V l (add by columns) and W j = V j (A λi)v j... (A λi) j 1 V j (add by rows) We next build a basis for E l (λ) in the following form. If V j {}, we start {u j,1,..., u j,qj } basis for V j. We then create sequences of j vectors for each of the above vectors. If u is any of them, we define v 1 = (A λi) j 1 u = (A λi)v 2 v 2 = (A λi) j 2 u = (A λi)v 3 (A λi) j 1 V j (A λi) j 2 V j so that We thus obtain a basis for. v j 1 = (A λi)u = (A λi)v j (A λi)vj v j = u, Vj, Av 1 = λv 1, Av k = v k 1 + λv k, k = 2,..., j. W j = (A λi) j 1 V j... (A λi)v j V j. If we change basis to this new basis (collect the corresponding basis for each row of the large decomposition above, noting that some rows might be zero), we have a Jordan matrix with q j blocks of size j j for each j. 13

Theorem 4. Every matrix is similar to a Jordan matrix (its Jordan canonical form). The Jordan form is unique up to permutation of its blocks, and it is the only general Jordan matrix such that the dimensions of the iterated kernels for all the eigenvalues coincide with those of A. Proof. Use the First Jordan Theorem to separate eigenvalues, and then use the construction of the second theorem. Corollary 1. The similarity class of a matrix is determined by the characteristic polynomials and the dimensions of all the iterated kernels (for all roots of the characteristic polynomial). Matrices with different Jordan forms are not similar. 7 Additional topics 7.1 Cayley Hamilton s Theorem Theorem 5. For every matrix χ A (A) =. Proof. A simple form of proving this theorem is using the Jordan form. Note that there are other proofs using much less complicated results. Let J be the Jordan form for A. Since χ J = χ A, we only need to prove that χ J (J) =. Hoever χ J (J k1 (λ 1 )) χ J (J k2 (λ 2 )) χ J (J) =... χ J (J kr(λ r )) and on the other hand which proves the theorem. (J k (λ) λi) k =, Another proof. It is possible to prove this same result using the First Jordan decomposition theorem. Using this theorem and the fact that p(a 1 ) p(p 1 p(a 2 ) AP ) =..., p(a k ) it is enough to prove the result for matrices with only one eigenvalue. Now, if the characteristic polynomial of A is (x λ) n, there exists l n such that E l (λ) = ker, (A λi) l = C n, which implies that (A λi) l = with l n. 14

Some comments. The minimal polynomial for A is the lowest degree monic polynomial satisfying Q(A) =. Assume that χ A (x) = (λ 1 x) m 1... (λ k x) m k. Since the minimal polynomial for A and its Jordan form J are the same, it is easy to see that if for every λ j we pick the minimum l j such that then the minimal polynomial is dim E lj (λ j ) = m j, (x λ 1 ) l 1... (x λ k ) l k. Another simple argument shows then that the following statements are equivalent: (a) the matrix is diagonalizable (b) the Jordan form of the matrix is diagonal (all blocks are 1 1) (c) all iterated kernels have maximum dimension E 2 (λ) = E 1 (λ), λ eigenvalue of A. (d) the minimal polynomial has simple roots. 7.2 The transposed Jordan form In many contexts it is common to meet a Jordan form made up of blocks of the form λ 1 λ......, 1 λ that is, with the 1s under the main diagonal. If we have been able to get to the usual Jordan form, we only need to reoder the basis to change to this transposed form, namely, if the vectors u 1, u 2,..., u p correspond to a p p block associated to λ, then the vectors u p, u p 1,..., u 1 give the same block with the one elements under the main diagonal. Corollary 2. Every matrix is similar to its transpose. Proof. A simple reordering of the basis shows that every Jordan matrix is similar to its transpose. Also, the tranpose of a matrix is similar to the transpose of its Jordan form. This finishes the proof. 15

7.3 The real Jordan form If A M(n; R), λ = α + βı (with β ) is an eigenvalue and we have the basis for E l (λ) giving the Jordan blocks for λ, then we just need to conjugate the vectors of the basis to obtain the basis giving the Jordan blocks for λ = α βı. The reason is simple: u E j (λ) u E j (λ) This means that the configuration of Jordan blocks for λ and λ are the same. If u 1,..., u p are the basis vectors associated to λ = α + βı (with β ), for a basis leading to Jordan form, then u 1,..., u p can be taken as the vectors associated to λ. We can then take v 1 = Re u 1, v 2 = Im u 1 v 3 = Re u 2, v 4 = Im u 2. v 2p 1 = Re u p, v 2p = Im u p (Re and Im denote the real and imaginary part of the vector). With these vectors, the blocks corresponding to λ and λ mix up, producing bigger blocks with real numbers. For instance [ α + βı 1 α β 1 α + βı [ β α 1 α β α βı 1 β α α βı In this way, when a real matrix has complex eigenvalues, we can restrict our bases to have real components and work with block-based Jordan blocks Λ I 2 Λ... [ α β J k (λ, λ) :=... I 2, Λ := β α Λ 7.4 Symmetric matrices Theorem 6. Every real symmetric matrix is diagonalizable. [ 1, I 2 := 1 Proof. Asssume that A M(n; R) satisfyes A T r = A. Let Au = λu, with u C n and λ C. Then λ u 2 = u T r A u = (A u) T r u = λ u 2, 16

which proves that λ is real. We now only need to prove that E 2 (λ) = E 1 (λ). This is a consequence of the following chain of implications: u E 2 (λ) = (A λi) 2 u = = u T r (A λi) 2 u = = ((A λi)u) T r ((A λi)u) = = (A λi)u 2 = = u E 1 (λ). 8 Exercises 1. (Section 1) Show that if BA = AB, then ker B and range B are A invariant. 2. (Section 1) Let V be A invariant and let W be such that F n = V W. Show that A is similar to a block upper triangular matrix. (Hint. Take a basis of F n composed of vectors of V followed by vectors of W.) 3. (Section 1) Let A be such that P 1 AP = 2 1 2 1 2 1 2 and let {p 1,..., p 4 } be the columns of P. span[p 1, p 2, p 3 are A invariant. Show that span[p 1, span[p 1, p 2 and 4. (Section 2) We can define the iterated ranges as follows with R (λ) = range(i) = C n. Show that (a) R j+1 (λ) R j (λ) for all j. R j (λ) = range(a λi) j, j 1 (b) If R l+1 (λ) = R l (λ), then R l+2 (λ) = R l+1 (λ). (Hint. This result can be proved directly or using the equivalent result for the iterated kernels.) (c) dim R j+1 (λ) dim R j (λ) dim R j (λ) dim R j 1 (λ). If R l+1 (λ) = R l (λ), what is the dimension of this subspace? 5. (Section 4) Let A be such that P 1 AP = J n (λ) and let {p 1, p 2,..., p n } be the column vectors of P. 17

(a) Show that p j E j (λ). (b) Show that (A λi) j 1 p j = p 1. (c) Use the previous results to find the minimal polynomial of p j. (d) Use (c) to find the minimal polynomial of A. 6. (Sections 5 and 6) In a 12-dimensional space, we let q 1 = 2, q 2 =, q 3 = 2, q 4 = 1, q j = j 5. (This corresponds to a Jordan matrix with two 1 1 blocks, two 3 3 blocks and one 4 4 blocks.) What is the associated solution of Problem (P)? In other words, what are the dimensions of the iterated kernels E j (λ) for a matrix with the Jordan structure given above? 7. (Section 6) Let A be such that χ A (x) = (2 x) 3 (4 x) 2. List all possible Jordan forms for A. List the associated minimal polynomial. (Hint. It is easy to find the minimal polynomial for all the vectors in the basis that leads to the Jordan form. The l.c.m. of the minimal polynomials for the vectors in any basis is the minimal polynomial for the matrix.) 8. (Section 6) Let A be such that χ A (x) = (3 x) 4 (2 x) 2 and minp A (x) = (x 3) 2 (x 2) 2. How many possible Jordan forms can A have? 9. (Section 6) Let A be such that χ A (x) = (2 x) 6. Write down all possible Jordan forms for A. For each of them, list the dimensions of the iterated kernels and the minimal polynomial of the matrix. 1. (Section 7) Prove that every complex Hermitian matrix has only real eigenvalues and is diagonalizable. (Hint. Instead of transposing, use conjugate transposition in the proof of Theorem 6.) References [1 Hernández, Eugenio. Algebra and Geometry (Spanish), Addison-Wesley. [2 Kress, Rainer. Linear Integral Operators. Springer Verlag. [3 Katznelson, Yitzhak, and Katznelson, Yonatan R. A (Terse) Introduction to Linear Algebra. AMS Student Mathematical Library. by F. J. Sayas Mathematics, Ho!, October 22 Revised and translated November 213 18