So I have this matrix here: $$ \begin{matrix} 0 & 2 & 0 \\ 0 & -1 & 1 \\ 0 & 0 & 1 \\ \end{matrix} $$ and I need to find the eigenvectors. I have already found the three eigenvalues of $0$, $-1$, and $1$. So, for $\lambda = 0$, I get this matrix: $$ \begin{matrix} 0 & 2 & 0 \\ 0 & -1 & 1 \\ 0 & 0 & 1 \\ \end{matrix} $$ but it has a column of zeros. I know there should be a free variable in this case but which one is it. After getting the variables, what would the eigenvector(s) in this case be. Edit: added another similar case, this is the matrix after adding the eigenvalues: $$ \begin{matrix} 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{matrix} $$ In this case, what would the eigenvectors be.
$\endgroup$ 31 Answer
$\begingroup$A non-zero vector ${\bf x}$ is in the eigenspace of $$ {\bf A} := \begin{pmatrix} 0 & 2 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{pmatrix} $$ with eigenvalue $\lambda = 0$ if and only if ${\bf A}{\bf x} = \lambda{\bf x} = 0\cdot {\bf x} = {\bf 0}$. In your case, this amounts to $$ \begin{pmatrix} 0 & 2 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} x_1\\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 0\\0\\0 \end{pmatrix} $$ which happens if and only if $$ \begin{pmatrix} 2x_2 \\ -x_2\\ x_3 \end{pmatrix} = \begin{pmatrix} 0\\0\\0 \end{pmatrix} $$
Here you can solve for $x_2 = 0$ and $x_3 = 0$, but there is no dependence on $x_1$, so you find that $$ {\bf A}{\bf x} = 0\cdot{\bf x} \iff {\bf x} = \begin{pmatrix} x_1\\x_2\\x_3 \end{pmatrix}= \begin{pmatrix} x_1\\0\\0 \end{pmatrix} $$ where we see that $x_1$ is now playing the role of the "free variable"
$\endgroup$ 3
