Page 241 - 35Linear Algebra

13

Diagonalization

Given a linear transformation, it is highly desirable to write its matrix with
respect to a basis of eigenvectors.

13.1 Diagonalizability

Suppose we are lucky, and we have L: V → V , and the ordered basis B =
(v 1 , . . . , v n ) is a set of eigenvectors for L, with eigenvalues λ 1 , . . . , λ n . Then:

L(v 1 ) = λ 1 v 1
L(v 2 ) = λ 2 v 2
. . .

L(v n ) = λ n v n

As a result, the matrix of L in the basis of eigenvectors B is diagonal:

     
x 1 λ 1 x 1
x
x
 2  λ 2   2
     . ,

 
. . . .
L  . =  .
 .    .
x n λ n x n
B B
where all entries off the diagonal are zero.
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