Degenerate eigenvalues, finding second eigenvector and possibly using Gram Schmidt process. I am having some trouble finding a second eigenvector.
I want to use a specific simple matrix say this 3 by 3 matrix.
\begin{equation*}
A = 
\begin{pmatrix}
5 & 4 & 2 \\
4 & 5 & 2 \\
2 & 2 & 2
\end{pmatrix}
\end{equation*}
I have used a computer to find that the eigenvalues are
{10,1,1}
I also know that the eigenvectors are
{(2,2,1), (-1,0,2),(-1,1,0)}
I know how to use the Gram-Schmidt process to find an orthonormal basis for vectors. It is the trick we do by projecting and subtracting.
A hint from a text I am reading says that one can find the other eigenvector using the Gram-Shmidt process.
It is my understanding that the problem might reduce to just finding an orthogonal vector to the eigenvector. How does one do this using Gram-Shmidt process in this particular case where is not so simple to write down the answer by inspection?
Basically, to be specific, for $\lambda = 1$ given $A$ above, how does one proceed?
The text in describing a matrix \begin{equation*}
H = 
\begin{pmatrix}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0
\end{pmatrix}
\end{equation*}
with similar properties hints that
"
It is convenient to describe the degenerate eigenspace for λ = 1 by identifying two mutually orthogonal vectors that span it. We can pick the first vector by choosing arbitrary values of $C$ and $C_0$
(an obvious choice is to set one of these, say $C_0$
, to zero). Then, using the Gram-Schmidt process (or in this case by simple inspection), we find a second eigenvector
orthogonal to the first. Here, this leads to
"
I changed the matrix because the one in the example had properties that could be seen just by inspection.
 A: Here is what the hint is getting at. If we have a symmetric matrix, by the spectral theorem there exists an orthogonal basis of eigenvectors (if you haven't learned this yet, don't worry). However, computing the eigenvectors in the usual way, or using a computer to find them, may only give you a basis of eigenvectors, it may not be an orthogonal basis. Typically, you need to take the basis of vectors you end up with and use the Gram-Schmidt process to make it an orthogonal basis. So let's take your example.
The eigenvector for the $\lambda=10$ eigenspace is $v_1=(2,2,1).$ The eigenvectors for the $\lambda = 1$ eigenspace are $v_2=(-1,0,2)$ and $v_3=(-1,1,0)$. Note that $v_2$ and $v_3$ are both orthogonal to $v_1$ but are not orthogonal to each other. So our task is to take the basis of the $1$-eigenspace $\left\{(-1,0,2),(-1,1,0) \right\}$ and use Gram-Schmidt to make it orthogonal. That way, the third vector in the original basis will be orthogonal to the other two. Doing this (or alternatively, using the cross product) we get the basis $\left\{(2,2,1), (-1,0,2), (4,-5,2) \right\}$. We see that with this one, the last two vectors are orthogonal to each other.
