# Looking for orthogonal basis of eigenvectors using Gram Schmidt process

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.)

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?

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Use Gram-Schmidt separately for each eigenspace. – Henning Makholm Apr 22 '12 at 22:57

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.