Let
be a square matrix of order
and let
be the corresponding linear transformation. In this section, we ask the
question ``does there exist a basis
of
such that
the matrix of the linear transformation
is in the simplest possible
form."
We know that, the simplest form for a matrix is the identity matrix and the
diagonal matrix. In this section, we show that for a certain class of matrices
we can find a basis
such that
is a diagonal matrix,
consisting of the eigenvalues of
This is equivalent to saying that
is similar to a diagonal matrix. To show the above, we need the
following definition.
DEFINITION 6.2.1 (Matrix Diagonalisation)
A matrix
is said to be diagonalisable if there exists a
non-singular matrix
such that
is a diagonal
matrix.
Remark 6.2.2
Let
be an
diagonalisable matrix with eigenvalues
By definition,
is similar to a
diagonal matrix
Observe that
as similar matrices have the same set of eigenvalues and
the eigenvalues of a diagonal matrix are its
diagonal entries.
THEOREM 6.2.4
let
be an
matrix. Then
is diagonalisable if and only
if
has
linearly independent eigenvectors.
Proof.
Let
be diagonalisable. Then there
exist matrices
and
such that
Or equivalently,
Let
Then
implies that
Since
's are the columns of a non-singular matrix
they are
non-zero and so for
we get the eigenpairs
of
Since,
's are columns of the non-singular matrix
using Corollary
4.3.9, we get
are linearly independent.
Thus we have shown that if
is diagonalisable then
has
linearly
independent eigenvectors.
Conversely, suppose
has
linearly independent eigenvectors
with eigenvalues
Then
Let
Since
are linearly independent, by
Corollary 4.3.9,
is non-singular. Also,
Therefore the matrix
is
diagonalisable.
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COROLLARY 6.2.5
let
be an
matrix. Suppose that the eigenvalues of
are distinct. Then
is
diagonalisable.
Proof.
As
is an
matrix, it has
eigenvalues.
Since all the eigenvalues of
are distinct, by Corollary
6.1.17, the
eigenvectors
are linearly independent. Hence, by Theorem
6.2.4,
is diagonalisable.
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Proof.
As
is diagonalisable,
by Theorem
6.2.4,
has
linearly independent eigenvalues.
Also,
as
. Hence, for each eigenvalue
,
has exactly
linearly independent eigenvectors.
Thus, for each
,
the homogeneous linear system
has exactly
linearly independent
vectors in its solution set. Therefore,
.
Indeed
for
follows from a simple counting argument.
Now suppose that for each
.
Then for each
, we can choose
linearly independent eigenvectors.
Also by Corollary 6.1.17, the eigenvectors corresponding to distinct
eigenvalues are linearly independent. Hence
has
linearly
independent eigenvectors. Hence by Theorem 6.2.4,
is diagonalisable.
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EXAMPLE 6.2.7
- Let
Then
Hence,
has eigenvalues
It is easily seen that
and
are the only eigenpairs.
That is, the matrix
has exactly one eigenvector corresponding
to the repeated eigenvalue
Hence, by Theorem 6.2.4,
the matrix
is not diagonalisable.
- Let
Then
Hence,
has eigenvalues
It can be easily
verified that
and
correspond to the eigenvalue
and
corresponds to the eigenvalue
Note that the set
consisting of
eigenvectors corresponding to the eigenvalue
are not orthogonal.
This set can be replaced by the orthogonal set
which still consists of eigenvectors corresponding to the eigenvalue
as
.
Also, the set
forms a basis of
So, by Theorem 6.2.4, the matrix
is diagonalisable.
Also, if
is the corresponding unitary
matrix then
Observe that the matrix
is a symmetric matrix. In this case, the
eigenvectors are mutually orthogonal. In general, for any
real
symmetric matrix
there always exist
eigenvectors and they are mutually
orthogonal. This result will be proved later.
A K Lal
2007-09-12