12.2 Schur’s Theorem
Recall that for a linear transformation, L ∈ℒ
a finite dimensional inner product
space, it could be represented in the form
is an orthonormal basis. Of course different bases will yield different matrices,
Schur’s theorem gives the existence of a basis in an inner product space such that
Definition 12.2.1 Let L ∈ℒ
where V is a vector space. Then a subspace U of V is
L invariant if L
In what follows, F will be the field of scalars, usually ℂ but maybe ℝ.
Theorem 12.2.2 Let L ∈ℒ
for H a finite dimensional inner product space
such that the restriction of L∗to every L invariant subspace has its eigenvalues in F.
Then there exist constants, cij for i ≤ j and an orthonormal basis,
The constants, cii are the eigenvalues of L. Thus the matrix whose ijth entry is cij is upper
Proof: If dim
for some k.
because by definition, w ⊗ w
Therefore, the theorem holds if H
is 1 dimensional.
Now suppose the theorem holds for n − 1 = dim
be an eigenvector for L∗
Dividing by its length, it can be assumed
. Using the Gram
Schmidt process, there exists an orthonormal basis for H
of the form
Denote by L1 the restriction of L to H1. Since H1 has dimension n− 1, the induction hypothesis
yields an orthonormal basis,
is an orthonormal basis for
because every vector in
has the property that its inner product with wn is 0 so in particular, this is true for the vectors
. Now define
to be the scalars satisfying
Then by 12.4,
If 1 ≤ k ≤ n − 1,
while from 12.3,
Since L = B on the basis
it follows L
It remains to verify the constants, ckk are the eigenvalues of L, solutions of the equation,
However, the definition of det
is the same as
where C is the upper triangular matrix which has cij for i ≤ j and zeros elsewhere. This equals 0
if and only if λ is one of the diagonal entries, one of the ckk. ■
Now with the above Schur’s theorem, the following diagonalization theorem comes very easily.
Recall the following definition.
Definition 12.2.3 Let L ∈ℒ
where H is a finite dimensional inner product space.
Then L is Hermitian if L∗
Theorem 12.2.4 Let L ∈ℒ
where H is an n dimensional inner product space. If L is
Hermitian, then all of its eigenvalues λk are real and there exists an orthonormal basis of
Proof: By Schur’s theorem, Theorem 12.2.2, there exist lij ∈ F such that
Then by Lemma 11.4.2,
By independence, if i
= j, lii
and so these are all real. If i < j,
it follows from independence
again that lij
= 0 because the coefficients corresponding to i < j
are all 0 on the right side.
Similarly if i > j,
it follows lij
= 0. Letting λk
That each of these wk is an eigenvector corresponding to λk is obvious from the definition of the
tensor product. ■