Page 242 - 35Linear Algebra

242 Diagonalization

Suppose that V is any n-dimensional vector space. We call a linear trans-
formation L: V 7→ V diagonalizable if there exists a collection of n linearly
independent eigenvectors for L. In other words, L is diagonalizable if there
exists a basis for V of eigenvectors for L.
In a basis of eigenvectors, the matrix of a linear transformation is diag-
onal. On the other hand, if an n × n matrix is diagonal, then the standard
basis vectors e i must already be a set of n linearly independent eigenvectors.
We have shown:

Theorem 13.1.1. Given an ordered basis B for a vector space V and a
linear transformation L: V → V , then the matrix for L in the basis B is
diagonal if and only if B consists of eigenvectors for L.

Non-diagonalizable example

Reading homework: problem 1

Typically, however, we do not begin a problem with a basis of eigenvec-
tors, but rather have to compute these. Hence we need to know how to
change from one basis to another:

13.2 Change of Basis

0
0
0
Suppose we have two ordered bases S = (v 1 , . . . , v n ) and S = (v , . . . , v )
1 n
0
for a vector space V . (Here v i and v are vectors, not components of vectors
i
0
in a basis!) Then we may write each v uniquely as
k
X
0
i
v = v i p ,
k
k
i
0
this is v as a linear combination of the v i . In matrix notation
k
p p · · · p
 1 1 1 
1 2 n
p 2 1 p 2 2 
0
0
v , v , · · · , v 0   .

1 2 n = v 1 , v 2 , · · · , v n  . .
 . . . . 
p n · · · p n
1 n
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