262 CHAPTER 12. SPECTRAL THEORY

Theorem 12.1.24 An n×n matrix is diagonalizable if and only if Fn has a basis of eigen-vectors of A. Furthermore, you can take the matrix S described above, to be given as

S =(

s1 s2 · · · sn

)where here the sk are the eigenvectors in the basis for Fn. If A is diagonalizable, theeigenvalues of A are the diagonal entries of the diagonal matrix.

Proof: To say that A is diagonalizable, is to say that

S−1AS =

λ 1

. . .

λ n

the λ i being elements of F. This is to say that for S =

(s1 · · · sn

), sk being the kth

column,

A(

s1 · · · sn

)=(

s1 · · · sn

)λ 1

. . .

λ n

which is equivalent, from the way we multiply matrices, that(

As1 · · · Asn

)=(

λ 1s1 · · · λ nsn

)which is equivalent to saying that the columns of S are eigenvectors and the diagonal matrixhas the eigenvectors down the main diagonal. Since S−1 is invertible, these eigenvectorsare a basis. Similarly, if there is a basis of eigenvectors, one can take them as the columnsof S and reverse the above steps, finally concluding that A is diagonalizable. ■

Example 12.1.25 Let A =

 2 0 01 4 −1−2 −4 4

 . Find a matrix, S such that

S−1AS = D,

where D is a diagonal matrix.

Solving det(λ I−A) = 0 yields the eigenvalues are 2 and 6 with 2 an eigenvalue of mul-tiplicity two. Solving (2I−A)x = 0 to find the eigenvectors, you find that the eigenvectorsare

a

 −210

+b

 101

where a,b are scalars. An eigenvector for λ = 6 is

 01−2

 . Let the matrix S be

S =

 −2 1 01 0 10 1 −2

