For matrix the matrix
$$A = \begin{bmatrix}
3&1&1\\
1&3&1\\
1&1&3\\
\end{bmatrix}$$
with eigenvalues $\lambda_1=5$, $\lambda_2=2$, $\lambda_3=2$, I am trying to find the corresponding eigenvector corresponding to the eigenvalue 2. I got
$$(A – 2I_3) = \begin{bmatrix}
1&1&1\\
1&1&1\\
1&1&1\\
\end{bmatrix}$$
Reducing it (row reduced echelon form), I get:
$$\left[
\begin{array}
{ccc|c}
1&1&1&0\\
0&0&0&0\\
0&0&0&0\\
\end{array}\right]$$
Ending up with $x_1 + x_2 + x_3 = 0$. How would I find the eigenvector from there? Usually, I end up getting two equations and it's easy from there. How would you do it with one?
Best Answer
For any square matrix with one value on the diagonal and another value everywhere else, a consistent pattern of (orthogonal) eigenvectors for the $n$ by $n$ case can be read from the columns of $$ \left( \begin{array}{rrrrrrrrrr} 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 2 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 3 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 4 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 5 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 6 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 7 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 8 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 9 \end{array} \right). $$ In your case, the upper left 3 by 3 corner. If you want the result orthonormal you need to divide each column by a square root of something appropriate. I have displayed the 10 by 10 version, notice how the diagonal numbers go up to 9 = 10 - 1.