Page 285 - A Course in Linear Algebra with Applications

8.1: Basic Theory of Eigenvectors 269

Since Xi,..., Xi are linearly independent, all the coefficients
(CJ —Ci +i)dj must vanish. But ci,..., q + 1 are all different, so
we can conclude that dj = 0 for j = 1,..., i; hence di +iX i+i =
0 and so di +i = 0, in contradiction to the original assumption.
Therefore the statement of the theorem must be correct.

The criterion for diagonalizability can now be established.

Theorem 8.1.6
Let A be an n x n matrix over a field F. Then A is diagonal-
izable if and only if A has n linearly independent eigenvectors
n
in F .
Proof
First of all suppose that A has n linearly independent eigen-
n
vectors in F , say Xi,..., X n, and that the associated eigen-
c
values are i,..., c n . Define S to be the n x n matrix whose
columns are the eigenvectors; thus

S=(X 1...X n).

The first thing to notice is that S is invertible; for by 8.1.5 its
columns are linearly independent. Forming the product of A
and S in partitioned form, we find that

AS = {AX X... AX n) = (c 1X 1 • • • c nX n),

which equals

'ci 0 0
0 c 2 0 0
(Xi ••• X n) = SD,

0 0 •

where D is the diagonal matrix with entries C\,..., c n. There-
1
fore S~ AS = D and A is diagonalizable.