Page 145 - Linear Algebra Done Right

The Spectral Theorem

(i, 1) (−i, 1)
,
√
√
2
2
2
is an orthonormal basis of C consisting of eigenvectors of T and that 133
with respect to this basis, the matrix of T is the diagonal matrix

2 + 3i 0
0 2 − 3i .
7.9 Complex Spectral Theorem: Suppose that V is a complex Because every
inner-product space and T ∈L(V). Then V has an orthonormal basis self-adjoint operator is
consisting of eigenvectors of T if and only if T is normal. normal, the complex
spectral theorem
Proof: First suppose that V has an orthonormal basis consisting of implies that every
eigenvectors of T. With respect to this basis, T has a diagonal matrix. self-adjoint operator on
The matrix of T ∗ (with respect to the same basis) is obtained by taking a finite-dimensional
the conjugate transpose of the matrix of T; hence T ∗ also has a diag- complex inner-product
onal matrix. Any two diagonal matrices commute; thus T commutes space has a diagonal
with T , which means that T must be normal, as desired. matrix with respect to
∗
To prove the other direction, now suppose that T is normal. There some orthonormal
is an orthonormal basis (e 1 ,...,e n ) of V with respect to which T has basis.
an upper-triangular matrix (by 6.28). Thus we can write
 
a 1,1 ... a 1,n
 . . 
7.10 M T, (e 1 ,...,e n ) =  . . . .  .
 
0 a n,n
We will show that this matrix is actually a diagonal matrix, which means
that (e 1 ,...,e n ) is an orthonormal basis of V consisting of eigenvectors
of T.
We see from the matrix above that

2 2
Te 1 =|a 1,1 |
and
2
2
2
2
T e 1 =|a 1,1 | +|a 1,2 | + ··· + |a 1,n | .
∗
Because T is normal, Te 1 = T e 1 (see 7.6). Thus the two equations
∗
above imply that all entries in the first row of the matrix in 7.10, except
possibly the first entry a 1,1 , equal 0.
Now from 7.10 we see that
2 2
Te 2 =|a 2,2 |