[Math] Finding eigenvectors when one column is zero

eigenvalues-eigenvectorslinear algebra

I have the matrix: $\begin{pmatrix}
3 & a & 0 \\
0 & 4 & 1 \\
0 & 2 & 5 \\
\end{pmatrix}$

and I should find every eigenvector for every eigenvalue.

After calculations I get that my eigenvalues are: $\lambda_1 = 6 $ and $ \lambda_2 = 3 = \lambda_3 $.

The eigenvector for $\lambda_1$ is: $t\begin{pmatrix}
a\\3\\6
\end{pmatrix}$.

However, for $\lambda = 3$ I get (after row reduction): $\begin{pmatrix}
0 & a & 0 \\
0 & 1 & 1 \\
0 & 0 & 0 \\
\end{pmatrix}$ which gives me:

$ax_2 = 0$ and $x_2 = – x_3$

The problem is that I don't really know where to go from here. I mainly have 2 questions:

  1. How do I handle the variable a?

  2. What do I do with $x_1 = 0$ ? (Especially unsure about this in general when it comes to eigenvectors)

Best Answer

Assuming your prior calculations are correct, the situation hinges on whether $a$ is zero or not. If $a \neq 0$, then $x_2$ is forced to be zero, so $x_3$ is as well, and $x_1$ is free. If $a=0$, then one of $x_2$ and $x_3$ can be made free, while $x_1$ is again free.

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