296 CHAPTER 12. SELF ADJOINT OPERATORS

such that (A 0

0 B

)(S11 S12

S21 S22

)

=

(S11 S12

S21 S22

)(D1 0

0 D2

)

Hence by block multiplication

AS11 = S11D1, BS22 = S22D2

BS21 = S21D1, AS12 = S12D2

It follows each of the xi is an eigenvector of A or else is the zero vector and that each of theyi is an eigenvector of B or is the zero vector. If there are n linearly independent xi, thenA is diagonalizable by Theorem 8.3.12 on Page 8.3.12.

The row rank of the matrix (x1, · · · ,xn+m) must be n because if this is not so, the rankof S would be less than n+m which would mean S−1 does not exist. Therefore, since thecolumn rank equals the row rank, this matrix has column rank equal to n and this meansthere are n linearly independent eigenvectors of A implying that A is diagonalizable. Similarreasoning applies to B. ■

The following corollary follows from the same type of argument as the above.

Corollary 12.1.4 Let Ak be an nk × nk matrix and let C denote the block diagonal(r∑

k=1

nk

)×

(r∑

k=1

nk

)

matrix given below.

C ≡

A1 0

. . .

0 Ar

 .

Then C is diagonalizable if and only if each Ak is diagonalizable.

Definition 12.1.5 A set, F of n×n matrices is said to be simultaneously diagonalizable ifand only if there exists a single invertible matrix S such that for every A ∈ F , S−1AS = DA

where DA is a diagonal matrix. F is a commuting family of matrices if whenever A,B ∈ F ,AB = BA.

Lemma 12.1.6 If F is a set of n×n matrices which is simultaneously diagonalizable, thenF is a commuting family of matrices.

Proof: Let A,B ∈ F and let S be a matrix which has the property that S−1AS is adiagonal matrix for all A ∈ F . Then S−1AS = DA and S−1BS = DB where DA and DB

are diagonal matrices. Since diagonal matrices commute,

AB = SDAS−1SDBS

−1 = SDADBS−1

= SDBDAS−1 = SDBS

−1SDAS−1 = BA.