Applied linear operators and spectral methods/Lecture 4

More on spectral decompositions edit

In the course of the previous lecture we essentially proved the following theorem:

Theorem: edit

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 of  .

Application edit

We can use the spectral representation to solve a system of linear homogeneous ordinary differential equations.

For example, we could wish to solve the system


(More generally   could be a   matrix.)

Comment: edit

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


Expanded out


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.

Theorem edit

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.

Theorem edit

If   is self-adjoint the following statements are true

  1.   is real for all  .
  2. All eigenvalues are real.
  3. Eigenvectors of distinct eigenvalues are orthogonal.
  4. There is an orthonormal basis formed by the eigenvectors.
  5. 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.

Linear manifold edit

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.

Invariant manifold edit

An invariant manifold   for the matrix   is the linear manifold for which   implies  .

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   If   then


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 that  .

We will explore some more of these ideas in the next lecture.

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