Jordan Canonical Form

Home , Hermitian matrix, Normal matrix, Symmetric matrix, Triangular matrix, Unitary matrix

• Jordan block

• Jordan canonical form • Extra material. Normal matrices

Deﬁnition. AJordanblockJk(λ)isak × k matrix with λ on the main diagonal and 1 above the main diagonal:   λo 100... 00    0 λo 10... 00    00λo 1 ... 00 λo   Jk =  000λo ... 00   ......    0000... λo 1 0000... 0 λo Properties of Jordan block: a) It has only one eigenvalue λ = λo with algebraic multiplicity k:   λo − λ 100... 00    0 λo − λ 10... 00    00λo − λ 1 ... 00 λo   Det(Jk )=Det 000λo − λ ... 00    ......    0000... λo − λ 1 0000... 0 λo − λ

k =(λo − λ) . b) Geometric multiplicity of λ = λo is 1:     0100... 00 1      0010... 00  0       0001... 00  0      Ker  0000... 00 =span 0  .     ......  ...     0000... 01 0 0000... 00 0 c) Denote       1 0 0        0   1   0         0   0   0        e1 =  0  ,e2 =  0  ,...,ek =  0  ,       ... ... ...       0 0 0 0 0 1

1 then λo λo Jk e1 = λoe1,Jk ei = λoei + ei−1,i=2, 3,...,k.

λo d) For B = Jk − λoIk we have

0=Be1,e1 = Be2,e2 = Be3,...,ek−1 = Bek.

Deﬁnition. A Jordan Canonical Form is a block-diagonal n × n matrix:   λ1 000 00 Jk1 ...  λ   0 J 2 00... 00 J =  k2  , ......  0000 0 λm ... Jkm

λ1 λ2 λm 0 with m Jordan blocks Jk1 , Jk2 , ..., Jkm , such that k1 + k2 + ...km = n,and denotes a zero matrix. Properties of the Jordan Canonical form

k1 k2 km a)det(J − λI)=(λ1 − λ) (λ2 − λ) ...(λm − λ) .

Proof Determinant of an upper triangular matrix. 1 1 1 b) The Jordan Canonical form has m eigenvectors x1,x2,...,xm. Each eigenvector corresponds to one Jordan block. Proof By induction on the number of Jordan blocks. Hint: show that all eigenvectors of J can be chosen to be orthogonal.

Theorem. Every n × n matrix A is similar to a Jordan Canonical Form:   λ1 000 00 Jk1 ...  λ2  −1  0 J 00... 00 S AS = J =  k2  . ......  0000 0 λm ... Jkm

1 1 1 Proof: Suppose x1,x2,...,xm are all linearly independent eigenvectors with (not necessarily distinct) eigenvalues λ1,λ2,...,λm:

j j Axk = λxk,j=1. (1)

2 3 2 3 If we construct exactly n linearly independent vectors x1,x1,..., x2,x2,..., ... such that

j j j−1 Axk = λxk + xk ,j>1. (2) then AS = SJ. We construct the vectors that satisfy (2) by induction on the dimension of the matrix. If the matrix is one-dimensional, the statement is obvious. Assume that all (n − r) × (n − r) matrices have Jordan form.

2 Since matrices A and A − µI have the same eigenvectors (Why?), without loss of generality, we can assume that one of the eigenvalues of A is zero (Why?), in other words A is singular. Therefore the image Im(A) has dimension rlinear map A from the image to the image of A:

A :Im(A) → Im(A). (3)

Denote by V =Im(A), and by Av the restriction of A on V .Then(3)is

Av : V → V.

Suppose dim(Ker(Av)=dim(Im(A)∩ Ker(A))= p. Then, there are p vectors xk ∈ V such that

j Axk = yk.

In other words we can add xk to the “ends” of chains in V . Dim(Ker(A)= n−r), hence there are n−r−p basis vectors zi, zi ⊥ Ker(A)∩Im(A), zi ∈ Ker(A). These z’s correspond to blocks J =(0). j We constructed vectors yk, xk and zk,thereareexactlyn of them, they satisfy (1) and (2). j Main issue in the proof: why all yk, xk and zk are linearly independent? Suppose j j ckyk + dkxk + gizi =0. j,k k i

Apply A,sinceAzi =0

 1    λkyk 0 j   j   j ck or + ck or + dkyk =0. j j−1 j−1 j,k,λ=0 j,k,λ=0 k,j=j ,λ=0 λkyk + yk yk max

j Since, by induction, all yk are linearly independent, all dk = 0. Hence j j ckyk = − gizi j,k i which is an equality between an element in the column space and an element in the space, orthogonal to column space. Hence the last equality is an equality of zeroes. Extra material: Normal matrices Recall that a (real-valued) matrix is symmetric if A = At. Deﬁnition. The conjugate transpose of a (complex-valued) matrix is

A∗ = A¯t.

In other words we take a transpose of a matrix and then take a complex conjugate of each of its entries. Question. What is the conjugate transpose of a (real-valued) symmetric matrix? Question. Recall that the (real-valued) symmetric matrices “commute” with respect to the (real-valued) inner product: x, Ay = Ax, y.

3 Deﬁne the complex-valued inner product as x, y = x¯iyi. Note that if x, y = y,x, then x, y is real.

How to define complex-valued matrices, that commute with the complex-valued inner product? Definition. A (complex-valued) matrix is Hermitian if A∗ = A. Hermitian matrices is a complex analog of symmetric matrices. Observation. Every eigenvalue of a hermitian matrix is real. Indeed, x, Ax is real, hence for any eigenvector x x, Ax = λ||x||2, and λ is real. Definition. A (complex-valued) matrix A is normal if A∗A = AA∗, that is, it commutes with its conjugate transpose. Definition. A (complex-valued) matrix A is unitary if A−1 = A∗, Lemma If a matrix A is normal, then it is diagonalizable. Note: Normal matrices are exactly those, that possess a complete set of orthonormal eigenvectors. Observe, that here we get even more, not only diagonalizability, but also orthogonality of eigenvectors. Proof: Step 1. For any square matrix A there is a unitary matrix U such that

U −1AU = T, where T is upper triangular with non-unity elements on the diagonal. Indeed, for any matrix A there is at least one eigenvector/eigenvalue:

Ax1 = λ1x1.

Let x1,x2,...,xn be an orthonormal basis, where x1 is the eigenvector and x2,x3,...,xn are arbitrary. Then   λ1 ∗∗∗... ∗∗    0 ∗∗∗... ∗∗    0 ∗∗∗... ∗∗ λ1 ∗ AU1 = U1M1,M1 =   =  0 ∗∗∗... ∗∗ 0 A1 ......  0 ∗∗∗... ∗∗

4 Continue inductively for A1,A2,...,An−1,and 10 U2 = 0 M2

Step 2.IfA is normal then T = U −1AU is also normal (why?), but a triangular normal matrix must be diagonal. Why? because the length of every row must equal the length of every column:

||Tx|| = ||T ∗x||.

Corollary. Symmetric and hermitian, orthogonal and unitary matrices are diagonalizable. Proof: They are normal. Question. For hermitian, orthogonal and unitary matrices

A = UDU−1 where U is complex-valued in general. Can we guarantee that U is always real-valued for symmetric matrices? Question. Yes, because all its eigenvectors can be chosen to be real. Why?