11.5. DIAGONALIZATION 197

Proof: First note that if S is n×n and its columns are linearly independent, then thesecolumns must also span all of Fn since otherwise, there would be a vector v not in thespan and you could add it in to the list and get n+1 vectors in an independent set. This iscontrary to Theorem 9.1.6. Thus Im(S) = Fn. Also, if Sx= Sy, then S (x−y) = 0 and sox= y (S is one to one.). Thus we can define S−1y 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 ∈ Fn, S(S−1x

)=(

SS−1)x= x. Hence SS−1 = I. By Problem 19 on Page 168, S−1S = I also. Alternatively,

S(S−1S

)=(SS−1

)S = S. Hence for all x,S

(S−1S

)x= Sx and so S

(S−1Sx− Ix

)= 0 and

so, since S is one to one, S−1Sx= Ix for all x showing that S−1S = 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 bothsides by S−1 and obtain x= 0 and so the columns of S are linearly independent.

Theorem 11.5.3 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. ■