298 CHAPTER 12. SELF ADJOINT OPERATORS

Now denote by F̃ the collection of matrices,{S−1CS : C ∈ F

}. Note F̃ features the single

matrix S.It follows easily that F̃ is also a commuting family of diagonalizable matrices. By Lemma

12.1.7 every B ∈ F̃ is a block diagonal matrix of the form given in 12.2 because each of thesecommutes with D described above as S−1AS and so by block multiplication, the diagonalblocks Bi corresponding to different B ∈ F̃ commute.

By Corollary 12.1.4 each of these blocks is diagonalizable. This is because B is known tobe so. Therefore, by induction, since all the blocks are no larger than n−1×n−1 thanks tothe assumption that A has more than one eigenvalue, there exist invertible ni×ni matrices,Ti such that T−1

i BiTi is a diagonal matrix whenever Bi is one of the matrices making up

the block diagonal of any B ∈ F̃ . It follows that for T defined by

T ≡

T1 0 · · · 0

0 T2. . .

......

. . .. . . 0

0 · · · 0 Tr

 ,

then T−1BT = a diagonal matrix for every B ∈ F̃ including D. Consider ST. It followsthat for all C ∈ F ,

T−1

something in F̃︷ ︸︸ ︷S−1CS T = (ST )

−1C (ST ) = a diagonal matrix. ■

Theorem 12.1.9 Let F denote a family of matrices which are diagonalizable. Then F issimultaneously diagonalizable if and only if F is a commuting family.

Proof: If F is a commuting family, it follows from Lemma 12.1.8 that it is simultaneouslydiagonalizable. If it is simultaneously diagonalizable, then it follows from Lemma 12.1.6 thatit is a commuting family. ■

12.2 Schur’s Theorem

Recall that for a linear transformation, L ∈ L (V, V ) for V a finite dimensional inner productspace, it could be represented in the form

L =∑ij

lijvi ⊗ vj

where {v1, · · · ,vn} is an orthonormal basis. Of course different bases will yield differentmatrices, (lij) . Schur’s theorem gives the existence of a basis in an inner product space suchthat (lij) is particularly simple.

Definition 12.2.1 Let L ∈ L (V, V ) where V is a vector space. Then a subspace U of V isL invariant if L (U) ⊆ U.

In what follows, F will be the field of scalars, usually C but maybe R.

Theorem 12.2.2 Let L ∈ L (H,H) for H a finite dimensional inner product space suchthat the restriction of L∗to every L invariant subspace has its eigenvalues in F. Then thereexist constants, cij for i ≤ j and an orthonormal basis, {wi}ni=1 such that

L =

n∑j=1

j∑i=1

cijwi ⊗wj

298 CHAPTER 12. SELF ADJOINT OPERATORSNow denote by F the collection of matrices, {S TOS :CEF } . Note F features the singlematrix S. 7It follows easily that F is also a commuting family of diagonalizable matrices. By Lemma12.1.7 every B € F is a block diagonal matrix of the form given in 12.2 because each of thesecommutes with D described above as S' ~tAS and so by block multiplication, the diagonalblocks B; corresponding to different B € F commute.By Corollary 12.1.4 each of these blocks is diagonalizable. This is because B is known tobe so. Therefore, by induction, since all the blocks are no larger than n—1 x n—1 thanks tothe assumption that A has more than one eigenvalue, there exist invertible n; x n; matrices,T; such that T, BT; is a diagonal matrix whenever B; is one of the matrices making upthe block diagonal of any B € F. It follows that for T’ defined byT O -- OT:T= 0 ;: . - 00 -- O TT,then T~!BT = a diagonal matrix for every B € F including D. Consider ST. It followsthat for all C € Ff,something in FT' s-'cs T=(ST)~'C(ST) = a diagonal matrix.Theorem 12.1.9 Let F denote a family of matrices which are diagonalizable. Then F issimultaneously diagonalizable if and only if F is a commuting family.Proof: If F is a commuting family, it follows from Lemma 12.1.8 that it is simultaneouslydiagonalizable. If it is simultaneously diagonalizable, then it follows from Lemma 12.1.6 thatit is a commuting family.12.2 Schur’s TheoremRecall that for a linear transformation, L ¢ £(V,V) for V a finite dimensional inner productspace, it could be represented in the formajwhere {v1,---,Vn} is an orthonormal basis. Of course different bases will yield differentmatrices, (1;;) . Schur’s theorem gives the existence of a basis in an inner product space suchthat (d;;) is particularly simple.Definition 12.2.1 Let L €¢ L(V,V) where V is a vector space. Then a subspace U of V isL invariant if L(U) CU.In what follows, F will be the field of scalars, usually C but maybe R.Theorem 12.2.2 Let L € £L(H,H) for H a finite dimensional inner product space suchthat the restriction of L*to every L invariant subspace has its eigenvalues in F. Then thereexist constants, c,; for i <j and an orthonormal basis, {w;};_, such thatno ojj=l i=l