Page 148 - Linear Algebra Done Right

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As an illustration of the real spectral theorem, consider the self-
3
adjoint operator T on R whose matrix (with respect to the standard
basis) is Chapter 7. Operators on Inner-Product Spaces
 
14 −13 8
 
 −13 14 8  .
8 8 −7
You should verify that

(1, −1, 0) (1, 1, 1) (1, 1, −2)
√ , √ , √
2 3 6
3
is an orthonormal basis of R consisting of eigenvectors of T and that
with respect to this basis, the matrix of T is the diagonal matrix
 
27 0 0

 0 9 0 
 .
0 0 −15
Combining the complex spectral theorem and the real spectral the-
orem, we conclude that every self-adjoint operator on V has a diagonal
matrix with respect to some orthonormal basis. This statement, which
is the most useful part of the spectral theorem, holds regardless of
whether F = C or F = R.

7.13 Real Spectral Theorem: Suppose that V is a real inner-product
space and T ∈L(V). Then V has an orthonormal basis consisting of
eigenvectors of T if and only if T is self-adjoint.

Proof: First suppose that V has an orthonormal basis consisting of
eigenvectors of T. With respect to this basis, T has a diagonal matrix.
This matrix equals its conjugate transpose. Hence T = T ∗ and so T is
self-adjoint, as desired.
To prove the other direction, now suppose that T is self-adjoint. We
will prove that V has an orthonormal basis consisting of eigenvectors
of T by induction on the dimension of V. To get started, note that our
desired result clearly holds if dim V = 1. Now assume that dim V> 1
and that the desired result holds on vector spaces of smaller dimen-
sion.
The idea of the proof is to take any eigenvector u of T with norm 1,
then adjoin to it an orthonormal basis of eigenvectors of T| {u} ⊥. Now