Linear Algebra, Theory And Applications, 2012a

13.1. SIMULTANEOUS DIAGONALIZATION 309
Lemma 13.1.7 Let D be a diagonal matrix of the form
⎛
⎞
λ 1 I n1 0 ··· 0
.
D ≡
0 λ 2 I .. . n2 ⎜
⎝
.
. .. . ⎟
..
0 ⎠ , (13.1)
0 ··· 0 λ r I nr
where I ni denotes the n i × n i identity matrix and λ i ≠ λ j for i ≠ j and suppose B is a
matrix which commutes with D. Then B is a block diagonal matrix of the form
⎛
⎞
B 1 0 ··· 0
.
B =
0 B ..
. 2 ⎜
⎝
.
. .. . ⎟
(13.2)
..
0 ⎠
0 ··· 0 B r
where B i is an n i × n i matrix.
Proof: Let B =(B ij )whereB ii = B i a block matrix as above in (13.2).
⎛
⎞
B 11 B 12 ··· B 1r
. B 21 B .. 22 B2r
⎜
⎝
.
. .. . ..
. ⎟
⎠
B r1 B r2 ··· B rr
Then by block multiplication, since B is given to commute with D,
Therefore, if i ≠ j, B ij =0.
λ j B ij = λ i B ij
Lemma 13.1.8 Let F denote a commuting family of n × n matrices such that each A ∈F
is diagonalizable. Then F is simultaneously diagonalizable.
Proof: First note that if every matrix in F has only one eigenvalue, there is nothing to
prove. This is because for A such a matrix,
S −1 AS = λI
and so
A = λI
Thus all the matrices in F are diagonal matrices and you could pick any S to diagonalize
them all. Therefore, without loss of generality, assume some matrix in F has more than one
eigenvalue.
The significant part of the lemma is proved by induction on n. If n =1, there is nothing
to prove because all the 1 × 1 matrices are already diagonal matrices. Suppose then that
the theorem is true for all k ≤ n − 1wheren ≥ 2andletF be a commuting family of
diagonalizable n × n matrices. Pick A ∈Fwhich has more than one eigenvalue and let
S be an invertible matrix such that S −1 AS = D where D is of the form given in (13.1).
By permuting the columns of S there is no loss of generality in assuming D has this form.
Now denote by ˜F the collection of matrices, { S −1 CS : C ∈F } . Note ˜F features the single
matrix S.