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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.

Hadi Nounou
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    The first equation (with the $0$ included on the right side) implies $x_2=0$. The third similarly implies $x_3=0$. $x_1$ is not constrained (because it doesn't appear in any of the equations), so that's your free variable. One routine way to construct eigenvectors for a given eigenspace is to find all the free variables, then set all but one free variable equal to zero and set one of them equal to $1$. – Ian Nov 24 '16 at 18:29
  • Ok thank you very much, but I also encountered another question with a similar layout. When applying the eigenvalue on this matrix , the new matrix is has two rows of zeroes, and one column of zeros. I added it above, if you can, please answer for it too. – Hadi Nounou Nov 24 '16 at 18:48
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    See http://math.stackexchange.com/a/1521354/265466 for how to read a kernel basis from a rref matrix. It doesn’t cover your case—no pivot in the first column—explicitly, but the third method described there makes this a no-brainer. – amd Nov 24 '16 at 20:42

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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"

Tom
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  • Just a question, I added a similar case now in my edit. In this case, I get a vector with x2+x3,0,0. What would the free variable be in this case, – Hadi Nounou Nov 24 '16 at 18:59
  • There would be two free variables in the second case since the dimension of the eigenspace with eigenvalue $0$ is two. Note that $x_1$ can be anything, and you only have the relation $x_2 + x_3 = 0$, so you can solve for $x_2$ as a function of $x_3$, say, and arrive at the solution $(x_1, -x_3 ,x_3)^{T} = x_1 (1, 0, 0)^{T} + x_3 (0, -1, 1)^{T}$. – Tom Nov 24 '16 at 19:21
  • Ok thank you very much. – Hadi Nounou Nov 25 '16 at 06:41