for V a finite dimensional inner product
space, it could be represented in the form
∑
L = lijvi ⊗ vj
ij
where
{v1,⋅⋅⋅,vn}
is an orthonormal basis. Of course different bases will yield different matrices,
(lij)
. Schur’s theorem gives the existence of a basis in an inner product space such that
(lij)
is
particularly simple.
Definition 12.2.1Let L ∈ℒ
(V,V)
where V is a vector space. Then a subspace U of V isL invariantif L
(U)
⊆ U.
In what follows, F will be the field of scalars, usually ℂ but maybe ℝ.
Theorem 12.2.2Let L ∈ℒ
(H,H )
for H a finitedimensional inner product spacesuch that the restriction of L^{∗}to every L invariant subspace has its eigenvalues in F.Then there exist constants, c_{ij}for i ≤ j and an orthonormal basis,
{wi }
_{i=1}^{n}suchthat
∑n ∑j
L = cijwi ⊗ wj
j=1i=1
The constants, c_{ii}are the eigenvalues of L. Thus the matrix whose ij^{th}entry is c_{ij}is uppertriangular.
Proof:If dim
(H)
= 1, let H = span
(w )
where
|w|
= 1. Then Lw = kw for some k.
Then
L = kw ⊗ w
because by definition, w ⊗ w
(w)
= w. Therefore, the theorem holds if H is 1 dimensional.
Now suppose the theorem holds for n − 1 = dim
(H)
. Let w_{n} be an eigenvector for L^{∗}.
Dividing by its length, it can be assumed
|wn |
= 1. Say L^{∗}w_{n} = μw_{n}. Using the Gram
Schmidt process, there exists an orthonormal basis for H of the form
{v1,⋅⋅⋅,vn−1,wn }
.
Then
(Lvk,wn ) = (vk,L∗wn) = (vk,μwn ) = 0,
which shows
L : H1 ≡ span (v1,⋅⋅⋅,vn−1) → span (v1,⋅⋅⋅,vn−1).
Denote by L_{1} the restriction of L to H_{1}. Since H_{1} has dimension n− 1, the induction hypothesis
yields an orthonormal basis,
{w1,⋅⋅⋅,wn −1}
for H_{1} such that
n−∑ 1∑j
L1 = cijwi⊗wj. (12.3)
j=1 i=1
(12.3)
Then
{w1, ⋅⋅⋅,wn }
is an orthonormal basis for H because every vector in
span (v1,⋅⋅⋅,vn−1)
has the property that its inner product with w_{n} is 0 so in particular, this is true for the vectors
It remains to verify the constants, c_{kk} are the eigenvalues of L, solutions of the equation,
det
(λI − L)
= 0. However, the definition of det
(λI − L)
is the same as
det(λI − C )
where C is the upper triangular matrix which has c_{ij} for i ≤ j and zeros elsewhere. This equals 0
if and only if λ is one of the diagonal entries, one of the c_{kk}. ■
Now with the above Schur’s theorem, the following diagonalization theorem comes very easily.
Recall the following definition.
Definition 12.2.3Let L ∈ℒ
(H,H )
where H is a finitedimensional inner product space.Then L is Hermitian if L^{∗} = L.
Theorem 12.2.4Let L ∈ℒ
(H,H )
where H is an n dimensional inner product space. If L isHermitian, then all of its eigenvalues λ_{k}are real and there exists an orthonormal basis ofeigenvectors
{wk }
such that
∑
L = λkwk⊗wk.
k
Proof:By Schur’s theorem, Theorem 12.2.2, there exist l_{ij}∈ F such that
n j n j
∑ ∑ lijwi⊗wj = L = L∗ = ∑ ∑ (lijwi⊗wj )∗
j=1 i=1 j=1 i=1
n j n i
= ∑ ∑ lijwj⊗wi = ∑ ∑ ljiwi⊗wj
j=1i=1 i=1 j=1
By independence, if i = j, l_{ii} =l_{ii} and so these are all real. If i < j, it follows from independence
again that l_{ij} = 0 because the coefficients corresponding to i < j are all 0 on the right side.
Similarly if i > j, it follows l_{ij} = 0. Letting λ_{k} = l_{kk}, this shows
L = ∑ λkwk ⊗ wk
k
That each of these w_{k} is an eigenvector corresponding to λ_{k} is obvious from the definition of the
tensor product. ■