Page 327 - A Course in Linear Algebra with Applications

9.1: Symmetric and Hermitian Matrices 311

x
to A = UDU* because U* = U~ . Next we perform a di-
rect computation to show that A commutes with its complex
transpose:
AA* = UDU*UD*U* = UDD*U*,

and in the same way
A* A = UD*U*UDU* = UD*DU*.
But diagonal matrices always commute, so DD* = D*D. It
follows that AA* = A*A, so that A is normal.
It remains to show that if A is normal, then there is an
orthonormal basis of C n consisting entirely of eigenvectors of
A. From 9.1.2 we know that there is a unitary matrix U such
that U* AU = T is upper triangular. The next observation
is that T is also normal. This too is established by a direct
computation:
T*T = U*A*UU*AU = U*(A*A)U.

In the same way TT* = U*(AA*)U. Since A*A = AA*, it
follows that T*T = TT*.
Now equate the (1, 1) entries of T*T and TT*; this yields
the equation
2 2 2 2
|
|tii| = iii| + |£i2| + --' + |iin| ,
which implies that £12,..., t\ n are all zero. By looking at the
3
(2, 2), (3, ),..., (n, n) entries of T*T and TT*, we see that
all the other off-diagonal entries of T vanish too. Thus T is
actually a diagonal matrix.
Finally, since AU = UT, the columns of U are eigenvec-
tors of A, and they form an orthonormal basis of C n because
U is unitary. This completes the proof of the theorem.

The last theorem provides us with many examples of di-
agonalizable matrices: for example, complex matrices which
are unitary or hermitian are automatically normal, as are real
symmetric and real orthogonal matrices. Any matrix of these
types can therefore be diagonalized by a unitary matrix.