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The next theorem is gives the existence of the Jordan canonical form.

Theorem A.0.5 Let A be an n× n matrix having eigenvalues λ 1, · · · ,λ r where the mul-tiplicity of λ i as a zero of the characteristic polynomial equals mi. Then there exists aninvertible matrix S such that

S−1AS =

J (λ 1) 0

. . .

0 J (λ r)

 (1.2)

where J (λ k) is an mk×mk matrix of the formJk1 (λ k) 0

Jk2 (λ k). . .

0 Jkr (λ k)

 (1.3)

where k1 ≥ k2 ≥ ·· · ≥ kr ≥ 1 and ∑ri=1 ki = mk.

Proof: From Corollary 18.4.4, there exists S such that S−1AS is of the form

T ≡

T1 · · · 0...

. . ....

0 · · · Tr

where Tk is an upper triangular mk ×mk matrix having only λ k on the main diagonal.By Corollary A.0.4 There exist matrices, Sk such that S−1

k TkSk = J (λ k) where J (λ k) isdescribed in 1.3. Now let M be the block diagonal matrix given by

M =

S1 0

. . .

0 Sr

 .

It follows that M−1S−1ASM = M−1T M and this is of the desired form. ■What about the uniqueness of the Jordan canonical form? Obviously if you change the

order of the eigenvalues, you get a different Jordan canonical form but it turns out that ifthe order of the eigenvalues is the same, then the Jordan canonical form is unique. In fact,it is the same for any two similar matrices.

Theorem A.0.6 Let A and B be two similar matrices. Let JA and JB be Jordan forms of Aand B respectively, made up of the blocks JA (λ i) and JB (λ i) respectively. Then JA and JBare identical except possibly for the order of the J (λ i) where the λ i are defined above.

Proof: First note that for λ i an eigenvalue, the matrices JA (λ i) and JB (λ i) are both ofsize mi×mi because the two matrices A and B, being similar, have exactly the same char-acteristic equation and the size of a block equals the algebraic multiplicity of the eigen-value as a zero of the characteristic equation. It is only necessary to worry about the num-ber and size of the Jordan blocks making up JA (λ i) and JB (λ i) . Let the eigenvalues of