12.1. SIMULTANEOUS DIAGONALIZATION 297

Lemma 12.1.7 Let D be a diagonal matrix of the form

D ≡

λ1In1

0 · · · 0

0 λ2In2

. . ....

.... . .

. . . 0

0 · · · 0 λrInr

 , (12.1)

where Ini denotes the ni × ni identity matrix and λi ̸= λj for i ̸= j and suppose B is amatrix which commutes with D. Then B is a block diagonal matrix of the form

B =

B1 0 · · · 0

0 B2. . .

......

. . .. . . 0

0 · · · 0 Br

 (12.2)

where Bi is an ni × ni matrix.

Proof: Let B = (Bij) where Bii = Bi a block matrix as above in 12.2.B11 B12 · · · B1r

B21 B22. . . B2r

.... . .

. . ....

Br1 Br2 · · · Brr

Then by block multiplication, since B is given to commute with D,

λjBij = λiBij

Therefore, if i ̸= j, Bij = 0. ■

Lemma 12.1.8 Let F denote a commuting family of n× n matrices such that each A ∈ Fis diagonalizable. Then F is simultaneously diagonalizable.

Proof: First note that if every matrix in F has only one eigenvalue, there is nothing toprove. This is because for A such a matrix,

S−1AS = λI

and soA = λI

Thus all the matrices in F are diagonal matrices and you could pick any S to diagonalizethem all. Therefore, without loss of generality, assume some matrix in F has more than oneeigenvalue.

The significant part of the lemma is proved by induction on n. If n = 1, there is nothingto prove because all the 1 × 1 matrices are already diagonal matrices. Suppose then thatthe theorem is true for all k ≤ n − 1 where n ≥ 2 and let F be a commuting family ofdiagonalizable n × n matrices. Pick A ∈ F which has more than one eigenvalue and letS be an invertible matrix such that S−1AS = D where D is of the form given in 12.1.By permuting the columns of S there is no loss of generality in assuming D has this form.