Theorem 7.4.6 is a special case of something known as diagonalization.
Definition 7.5.1 An n × n matrix A is diagonalizable if there exists an invertible matrix S such
where D is a diagonal matrix.
The following theorem gives the condition under which a matrix is diagonalizable.
Theorem 7.5.2 An 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 Fn 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
Also, if Sx
and so x
is one to one.). Thus we can define S−1y
to be that
vector such that S
. Then S−1
is a linear transformation because of the following
Thus, since S
is one to one, as explained above, it follows that
Therefore, there is a matrix, still denoted as S−1 such that for any x ∈ Fn, S
By Problem 19
on Page 310
also. Alternatively, S
Hence for all x,S
and so S
and so, since S
is one to one, S−1Sx
showing that S−1S
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 7.5.3 An n × n matrix is diagonalizable if and only if Fn has a basis of eigenvectors of A.
Furthermore, you can take the matrix S described above, to be given as
where here the sk are the eigenvectors in the basis for Fn. If A is diagonalizable, the eigenvalues of A are
the diagonal entries of the diagonal matrix.
Proof: To say that A is diagonalizable, is to say that
the λi being elements of F. This is to say that for S =
being the kth
which is equivalent, from the way we multiply matrices, that
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. ■