More on spectral decompositions[edit | edit source]
In the course of the previous lecture we essentially proved the following theorem:
1) If a matrix has linearly independent real or complex
eigenvectors, the can be diagonalized.
2) If is a matrix whose columns are eigenvectors then
is the diagonal matrix of eigenvalues.
The factorization is called the spectral representation
We can use the spectral representation to solve a system of linear homogeneous ordinary
For example, we could wish to solve the system
(More generally could be a matrix.)
Higher order ordinary differential equations can be reduced to this form. For example,
Then the system of equations is
Returning to the original problem, let us find the eigenvalues and eigenvectors of
. The characteristic equation is
o we can calculate the eigenvalues as
The eigenvectors are given by
Possible choices of and are
The matrix is one whose columd are the eigenvectors of , i.e.,
If the system of equations becomes
The solutions of these equations are
This is the solution of the system of ODEs that we seek.
Most "generic" matrices have linearly independent eigenvectors. Generally a matrix will
have distinct eigenvalues unless there are symmetries that lead to repeated values.
If has distinct eigenvalues then it has linearly independent eigenvectors.
We prove this by induction.
Let be the eigenvector corresponding to the eigenvalue . Suppose
are linearly independent (note that this is true for
= 2). The question then becomes: Do there exist
not all zero such that the linear combination
Let us multiply the above by . Then, since
, we have
Since is arbitrary, the above is true only when
In thast case we must have
This leads to a contradiction.
Therefore are linearly independent.
Another important class of matrices which are diagonalizable are those which are self-adjoint.
If is self-adjoint the following statements are true
- is real for all .
- All eigenvalues are real.
- Eigenvectors of distinct eigenvalues are orthogonal.
- There is an orthonormal basis formed by the eigenvectors.
- The matrix can be diagonalized (this is a consequence of the previous statement.)
1) Because the matrix is self-adjoint we have
From the property of the inner product we have
which implies that is real.
2) Since is real, is real.
Also, from the eiegnevalue problem, we have
Therefore, is real.
3) If and are two eigenpairs then
Since the matrix is self-adjoint, we have
Therefore, if , we must have
Hence the eigenvectors are orthogonal.
4) This part is a bit more involved. We need to define a manifold first.
A linear manifold (or vector subspace) is a subset of which is
closed under scalar multiplication and vector addition.
Examples are a line through the origin of -dimensional space, a plane through the origin,
the whole space, the zero vector, etc.
An invariant manifold for the matrix is the linear manifold for which
Examples are the null space and range of a matrix . For the case of a rotation
about an axis through the origin in -space, invaraiant manifolds are the origin, the
plane perpendicular to the axis, the whole space, and the axis itself.
Therefore, if are a basis for and
are a basis for (the perpendicular component of
) then in this basis has the representation
We need a matrix of this form for it to be in an invariant manifold for .
Note that if is an invariant manifold of it does not follow that
is also an invariant manifold.
Now, if is self adjoint then the entries in the off-diagonal spots must be zero too.
In that case, is block diagonal in this basis.
Getting back to part (4), we know that there exists at least one eigenpair ()
(this is true for any matrix). We now use induction. Suppose that we have found ()
mutually orthogonal eigenvectors with and
are real, . Note that the s are invariant manifolds of as
is the space spanned by the s and so is the manifold perpendicular to these vectors).
We form the linear manifold
This is the orthogonal component of the eigenvectors
Therefore which means that is invariant.
Hence contains at least one eigenvector with real eigenvalue .
We can repeat the procedure to get a diagonal matrix in the lower block of the block
diagonal representation of . We then get distinct eigenvectors and so can
be diagonalized. This implies that the eigenvectors form an orthonormal basis.
5) This follows from the previous result because each eigenvector can be normalized so
We will explore some more of these ideas in the next lecture.