I have the matrix
$$I = ma^2\begin{bmatrix} 4 & -1 & -1\\ -1 & 4 & -1 \\ -1 & -1 & 4\end{bmatrix}$$
The eigenvalues are $\lambda _{1,2,3} = (5ma^2,5ma^2,2ma^2)$.
I need to generate an orthonormal set of vectors from the degenerate eigenvalues.
I looked up a way to do it but I don't understand it. The Gram-Schmidt procedure takes input vectors $v_i$ and outputs an orthonormal set of vectors $u_i$. The process can be summarized as follows:
$u_1 = v_1$
$u_2 = v_2 - $(projection of $v_2$ onto $v_1$)
$u_3 = v_3 - $(projection of $v_3$ onto $v_2$) $-$(projection of $v_3$ onto $v_1$)
Then we simply normalize the $u_i$ vectors. But what do I use for $v_1, v_2, $and $v_3$?
You can certainly use the generic procedure for finding eigenvalues and eigenvectors for this problem. Since this is a real symmetric matrix, you’ll end up with a full set of linearly independent eigenvectors $\{v_1,v_2,v_3\}$ and can then apply the Gram-Schmidt process to this set to produce an orthonormal set. The latter incrementally builds an orthonormal set out of the given vectors by throwing away the components of each successive vector that are in the directions of the orthonormal vectors generated up to that point (that’s what the projections onto the other vectors are). Because of the special form of your matrix, however, we can avoid all this.
Ignoring the constant factor of $ma^2$ for the moment, since its only effect will be to scale all of the eigenvalues of the matrix, observe that we can decompose your matrix into $5I-B$, where $B$ is the $3\times3$ matrix that consists entirely of $1$s. If $\mathbf v$ is an eigenvector of $B$ with eigenvalue $\lambda$, then $(5I-B)\mathbf v=(5-\lambda)\mathbf v$, so $\mathbf v$ is also an eigenvector of your matrix with eigenvalue $5-\lambda$. $B$ has rank one, so has zero as a double eigenvalue, and the trace of $B$ tells us that the remaining eigenvalue is $3$, so the eigenvalues of your matrix are $5$, $5$ and $2$.
We can find eigenvectors by computing kernels of various singular matrices derived from the eigenvalues as usual, but can also take a short cut here. The row sums of $B$ (and of your matrix) are identical, but summing over the rows is the same as right-multiplying by $(1,1,1)^T$, so that’s an eigenvector, and it corresponds to the eigenvalue $5-3=2$. We know that a real symmetric matrix is diagonalizable, so the eigenspace of $5$ is two-dimensional. In addition, eigenspaces of different eigenvalues of such a matrix are orthogonal, which means that we can take any linearly independent pair of vectors orthogonal to $(1,1,1)^T$ as our eigenvectors of $5$. To avoid having to go through the Gram-Schmidt process, we take advantage of the fact that we’re working in $\mathbb R^3$. First, we pick some vector that’s orthogonal to $(1,1,1)^T$ by inspection: $(1,-1,0)^T$ is an obvious choice. Now we want a vector that’s orthogonal to them both, so we compute the cross product $(1,1,1)^T\times(1,-1,0)^T=(1,1,-2)^T$. You can verify that these are indeed eigenvectors of $5$. All that’s left to do is normalize these three vectors. Physically, the fact that we can choose any two vectors orthogonal to the principal axis $(1,1,1)^T$ as our other principal axes reflects the symmetry of the masses about this axis.