9
$\begingroup$

Question

Find an orthogonal basis of eigenvectors for the following matrix. The matrix has a repeated eigenvalue so you will need to use the Gram-Schmidt process.

$$\begin{bmatrix}5 & 4 & 2\\ 4 & 5 & 2 \\ 2 & 2 & 2 \end{bmatrix}$$

($\lambda = 1$ is a double eigenvalue.)

Answer

Well here's what I found for eigenvalues and eigenvectors -

For $\lambda = 1$ I found eigenvectors $\begin{bmatrix}-1 \\ 1 \\ 0 \end{bmatrix}$, $\begin{bmatrix}-1 \\ 0 \\ 2 \end{bmatrix}$

For $\lambda = 10$ I found eigenvector $\begin{bmatrix}2 \\ 2 \\ 1 \end{bmatrix}$

Now I can see that the eigenvectors of $\lambda = 1$ are not orthogonal to each other. But how do I do the rest of the question? It says to use the Gram-Schmidt process but if I use that on one of the vectors it will change it's orientation and it will no longer be an eigenvector. So how do I get an orthogonal basis while getting to keep the basis as eigenvectors?

$\endgroup$
2
  • 9
    $\begingroup$ Use Gram-Schmidt separately for each eigenspace. $\endgroup$
    – hmakholm left over Monica
    Apr 22, 2012 at 22:57
  • 1
    $\begingroup$ Within an eigenspace corresponding to particular eigenvalue, GS will maintain the eigenvector property, since it only performs linear operations (projection). That is $$A(v_1+v_2) = \lambda (v_1 + v_2)$$. Then, we also have that eigenvectors corresponding to different eigenvalues are orthogonal and we are done $\endgroup$
    – information_interchange
    Mar 28, 2020 at 1:11

1 Answer 1

Reset to default
14
$\begingroup$

The Gram-Schmidt process does not change the span. Since the span of the two eigenvectors associated to $\lambda=1$ is precisely the eigenspace corresponding to $\lambda=1$, if you apply Gram-Schmidt to those two vectors you will obtain a pair of vectors that are orthonormal, and that span the eigenspace; in particular, they will also be eigenvectors associated to $\lambda=1$.

You can also see that the eigenvector corresponding to $\lambda=10$ is orthogonal to the other two eigenvectors, hence to the entire eigenspace they span. So the third eigenvector will already be orthogonal to the orthonormal basis you find for $E_{1}$. You'll just need to normalize it.

Note. You can always find an orthonormal basis for each eigenspace by using Gram-Schmidt on an arbitrary basis for the eigenspace (or for any subspace, for that matter). In general (that is, for arbitrary matrices that are diagonalizable) this will not produce an orthonormal basis of eigenvectors for the entire space; but since your matrix is symmetric, the eigenspaces are mutually orthogonal so the process will definitely work. In fact, a real matrix is orthogonally diagonalizable if and only if it is symmetric.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.