First of all, here is what it means for two matrices to be similar.
Definition A.7.11 Let A,B be two n×n matrices. Then they are similar if and only if there exists an
invertible matrix S such that
Proposition A.7.12 Define for n × n matrices A ∼ B if A is similar to B. Then
Proof: It is clear that A ∼ A because you could just take S = I. If A ∼ B, then for some S
which shows that B ∼ A.
Now suppose A ∼ B and B ∼ C. Then there exist invertible matrices S,T such that
showing that A is similar to C. ■
For your information, when ∼ satisfies the above conditions, it is called a similarity relation. Similarity
relations are very significant in mathematics.
When a matrix is similar to a diagonal matrix, the matrix is said to be diagonalizable. I think this is
one of the worst monstrosities for a word that I have ever seen. Nevertheless, it is commonly used
in linear algebra. It turns out to be the same as nondefective. The following is the precise
Definition A.7.13 Let A be an n × n matrix. Then A is diagonalizable if there exists an invertible
matrix S such that
where D is a diagonal matrix. This means D has a zero as every entry except for the main diagonal. More
precisely, Dij = 0 unless i = j. Such matrices look like the following.
where ∗ might not be zero.
The most important theorem about
is the following major result.
Theorem A.7.14 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 vk 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: Suppose there exists a basis of eigenvectors
Then let S
be given as
above. It follows S−1
exists because these vectors are linearly independent so N
is one to one which implies det
0 which implies S−1
exists. Let S−1
be of the
where wkTvj = δkj. Then
Next suppose A is diagonalizable so that S−1AS = D. Let
where the columns are the vk and
showing the vi are eigenvectors of A and the λk are eigenvectors. Now the vk form a basis for Fn because
the matrix S having these vectors as columns is given to be invertible. ■
In other words, to diagonalize A you get a basis of eigenvectors
a diagonal matrix which has the eigenvalues down the main diagonal listed according to
multiplicity. Note also that for n
a positive integer,
The interior S−1S cancel and so this reduces to
and it is easy to compute Dm. More generally, you can define functions of the matrix using power series in