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So I have this matrix here: $$ \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} $$ and I need to find the eigenvectors. I have already found the eigenvalue of $1$. So, for $\lambda = 1$, I get this matrix: $$ \begin{pmatrix} 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix} $$ 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.

Alex Silva
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Hadi Nounou
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  • See http://math.stackexchange.com/a/1521354/265466 for how to read a kernel basis from a rref matrix. – amd Nov 24 '16 at 20:40

1 Answers1

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The free variables are $x_1$ and $x_3$. This is because there is no leading one in columns one and three. So the eigenvectors are $\begin{pmatrix}1 \\ 0 \\ 0\end{pmatrix}$ and $\begin{pmatrix}0 \\ -1 \\ 1\end{pmatrix}$. These are the eigenvectors, because if you multiply the matrix $\begin{pmatrix}0&1&1 \\ 0&0&0 \\ 0&0&0\end{pmatrix}$ by either one of these vectors you get the zero vector.

Dave
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