Page 286 - A Course in Linear Algebra with Applications

270 Chapter Eight: Eigenvectors and Eigenvalues

Conversely, assume that A is diagonalizable and that
1
c
S~ AS = D is a diagonal matrix with entries i,..., c n . Then
AS = SD. This implies that if X{ is the zth column of S,
then AXi equals the ith column of SD, which is CjXj. Hence
Xi,..., X n are eigenvectors of A associated with eigenvalues
c\,..., c n . Since X\,..., X n are columns of the invertible ma-
trix S, they must be linearly independent. Consequently A
has n linearly independent eigenvectors.

Corollary 8.1.7
An n x n complex matrix which has n distinct eigenvalues is
diagonalizable.
This follows at once from 8.1.5 and 8.1.6. On the other
hand, it is easy to think of matrices which are not diagonaliz-
able: for example, there is the matrix

-(; o-

Indeed if A were diagonalizable, it would be similar to the
identity matrix I2 since both its eigenvalues equal 1, and
1
S~ AS = I2 for some S; but the last equation implies that
A = SI 2S~ l = I2, which is not true.
An interesting feature of the proof of 8.1.6 is that it pro-
vides us with a method of finding a matrix S which diagonal-
izes A. One has simply to find a set of linearly independent
eigenvectors of A; if there are enough of them, they can be
taken to form the columns of the matrix S.

Example 8.1.5

Find a matrix which diagonalizes A =

In Example 8.1.1 we found the eigenvalues of A to be
3 + A/^T and 3 — y/^T; hence A is diagonalizable by 8.1.7. We