Theorem 7.4.6 is a special case of something known as diagonalization.

Definition 7.5.1An n × n matrix A is diagonalizable if there exists an invertible matrix S suchthat

S− 1AS = D

where D is a diagonal matrix.

The following theorem gives the condition under which a matrix is diagonalizable.

Theorem 7.5.2An n×n matrix S is invertible if and only if its columns are linearly independent.

Proof:First note that if S is n×n and its columns are linearly independent, then these columns must
also span all of F^{n} since otherwise, there would be a vector v not in the span and you could add it in to the
list and get n + 1 vectors in an independent set. This is contrary to Theorem 6.1.6. Thus Im

(S)

= F^{n}.
Also, if Sx = Sy, then S

(x − y )

= 0 and so x = y (S is one to one.). Thus we can define S^{−1}y to be that
vector such that S

( )
S−1y

= y. Then S^{−1} is a linear transformation because of the following
reasoning.

( )
S S−1(ax +by) = ax+ by

( ) ( ) ( )
S aS−1x+ bS− 1y = aS S −1x + bS S −1y
= ax+ by

Thus, since S is one to one, as explained above, it follows that

S −1(ax+ by) = aS−1x+ bS− 1y

Therefore, there is a matrix, still denoted as S^{−1} such that for any x ∈ F^{n}, S

(S−1x)

=

(SS− 1)

x = x.
Hence SS^{−1} = I. By Problem 19 on Page 310, S^{−1}S = I also. Alternatively, S

( −1 )
S S

=

( −1)
SS

S = S.
Hence for all x,S

( −1 )
S S

x = Sx and so S

( −1 )
S Sx − Ix

= 0 and so, since S is one to one, S^{−1}Sx = Ix for
all x showing that S^{−1}S = I also. ■

Thus if the columns of a matrix are linearly independent, then the matrix is invertible. On the other
hand, if the matrix S is invertible, then if Sx = 0 one could multiply both sides by S^{−1} and obtain x = 0
and so the columns of S are linearly independent.

Theorem 7.5.3An n × n matrix is diagonalizable if and only if F^{n}has a basis of eigenvectors of A.Furthermore, you can take the matrix S described above, to be given as

( )
S = s1 s2 ⋅⋅⋅ sn

where here the s_{k}are the eigenvectors in the basis for F^{n}. If A is diagonalizable,the eigenvalues of A arethe diagonal entries of the diagonal matrix.

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

( )
| λ1 |
S−1AS = |( ... |)
λn

the λ_{i} being elements of F. This is to say that for S =

( )
s1 ⋅⋅⋅ sn

, s_{k} being the k^{th}
column,

( )
λ1
A ( s ⋅⋅⋅ s ) = ( s ⋅⋅⋅ s ) || .. ||
1 n 1 n ( . )
λn

which is equivalent, from the way we multiply matrices, that

( ) ( )
As1 ⋅⋅⋅ Asn = λ1s1 ⋅⋅⋅ λnsn

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