Page 320 - A Course in Linear Algebra with Applications

304 Chapter Nine: Advanced Topics

Theorem 9.1.1
Let A be a hermitian matrix. Then:
(a) the eigenvalues of A are all real;
(b) eigenvectors of A associated with distinct eigenvalues
are orthogonal.

Proof
Let c be an eigenvalue of A with associated eigenvector X, so
that AX — cX. Taking the complex transpose of both sides
of this equation and using 7.1.7, we obtain X*A = cX* since
A = A*. Now multiply both sides of this equation on the right
2
by X to get X*AX = cX*X = c||X|| : remember here that
X*X equals the square of the length of X. But (X*AX)* =
X*A*X** = X*AX; thus the scalar X* AX equals its complex
conjugate and so it is real. It follows that c||X|| 2 is real. Since
lengths of vectors are always real, we deduce that c, and hence
c, is real, which completes the proof of (a).
To prove (b) take two eigenvectors X and Y associated
with distinct eigenvalues c and d. Thus AX = cX and AY =
dY. Then Y*AX = Y*(cX) = cY*X, and in the same
way X*AY = dX*Y. However, by 7.1.7 again, (X*AY)* =
Y*A*X = Y*AX. Therefore (dX*Y)* = cY*X, or dY*X =
cY*X because d is real by the first part of the proof. This
means that (c—d)Y*X = 0, from which it follows that Y*X =
0 since c ^ d. Thus X and Y are orthogonal.

Suppose now that {Xi,..., X r} is a set of linearly inde-
pendent eigenvectors of the n x n hermitian matrix A, and
that r is chosen as large as possible. We can multiply Xi by
l/||Xj|| to produce a unit vector; thus we may assume that
each Xi is a unit vector. By 9.1.1 {Xi,..., X r} is an orthonor-
mal set. Now write U = (X\ .. -X r), an n x r matrix. Then
U has the property

AU = (AYi ... AX r) = {c 1X 1 ... c rX r),