Page 156 - Linear Algebra Done Right

Chapter 7. Operators on Inner-Product Spaces
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Proof: To prove the easy direction, first suppose that there is an
orthonormal basis of V such that the matrix of T is a block diagonal
matrix where each block is a 1-by-1 matrix or a 2-by-2 matrix of the
form 7.26. With respect to this basis, the matrix of T commutes with
the matrix of T ∗ (which is the conjugate of the matrix of T), as you
should verify (use formula 7.24 for the product of two block diagonal
matrices). Thus T commutes with T , which means that T is normal.
∗
To prove the other direction, now suppose that T is normal. We will
prove our desired result by induction on the dimension of V. To get
started, note that our desired result clearly holds if dim V = 1 (trivially)
or if dim V = 2 (if T is self-adjoint, use the real spectral theorem 7.13;
if T is not self-adjoint, use 7.15).
Now assume that dim V> 2 and that the desired result holds on
vector spaces of smaller dimension. Let U be a subspace of V of di-
mension 1 that is invariant under T if such a subspace exists (in other
words, if T has a nonzero eigenvector, let U be the span of this eigen-
vector). If no such subspace exists, let U be a subspace of V of dimen-
sion 2 that is invariant under T (an invariant subspace of dimension 1
or 2 always exists by 5.24).
In a real vector space If dim U = 1, choose a vector in U with norm 1; this vector will
with dimension 1, there be an orthonormal basis of U, and of course the matrix of T| U is a
are precisely two 1-by-1 matrix. If dim U = 2, then T| U is normal (by 7.18) but not self-
vectors with norm 1. adjoint (otherwise T| U , and hence T, would have a nonzero eigenvector;
see 7.12), and thus we can choose an orthonormal basis of U with re-
spect to which the matrix of T| U has the form 7.26 (see 7.15).
Now U ⊥ is invariant under T and T| U ⊥ is a normal operator on U ⊥
(see 7.18). Thus by our induction hypothesis, there is an orthonormal
basis of U ⊥ with respect to which the matrix of T| U ⊥ has the desired
form. Adjoining this basis to the basis of U gives an orthonormal basis
of V with respect to which the matrix of T has the desired form.

Positive Operators

Many mathematicians An operator T ∈L(V) is called positive if T is self-adjoint and
also use the term Tv, v ð 0
positive semidefinite
operator, which means for all v ∈ V. Note that if V is a complex vector space, then the
the same as positive condition that T be self-adjoint can be dropped from this definition
operator. (by 7.3).