Page 249 - Advanced Linear Algebra

Structure Theory for Normal Operators 233

)
2 Conversely, if
=~ : p Ä p :
and if is orthogonal projection onto : , then b Ä b ~ is an
orthogonal resolution of the identity.
Proof. To prove 1), if bÄb ~ is an orthogonal resolution of the
identity, Theorem 2.25 implies that
=~ im ² ³ l Ä l im ² ³

However, since the 's are pairwise orthogonal and self-adjoint, it follows that

º #Á $»~º#Á $»~º#Á »~
and so

=~ im ² ³ p Ä p im ² ³

For the converse, Theorem 2.25 implies that bÄb ~ is a resolution of

the identity where is projection onto im²³ along

ker²³ ~ im ² ³ ~ im ²³ ž

£
Hence, is orthogonal.

Unitary Diagonalizability
We have seen (Theorem 8.10) that a linear operator B ²= ³ on a finite-
dimensional vector space is diagonalizable if and only if
=
=~ ; l Ä l ;

Of course, each eigenspace ; has an orthonormal basis E , but the union of
these bases need not be an orthonormal basis for .
=
(
B
=

Definition A linear operator ²= ³ is unitarily diagonalizable when is
(
)
)
complex and orthogonally diagonalizable when = is real if there is an
ordered orthonormal basis E of for which the matrix ~²" Á Ã Á " ³ = ´ µ E is
diagonal, or equivalently, if
"~ "
for all ~ Á Ã Á .
Here is the counterpart of Theorem 8.10 for inner product spaces.
Theorem 10.7 Let = be a finite-dimensional inner product space and let
B ²= ³. The following are equivalent:
)
(
)
1 is unitarily orthogonally diagonalizable.
)
2 = has an orthonormal basis that consists entirely of eigenvectors of .