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Could anyone help in how to get the eigenvalue and eigenvectors of a matrix that contain both zero column and zero row like : \begin{pmatrix} -1 & 1 & 0\\ 1 & -1 & 0\\ 0 & 0 & 0\\ \end{pmatrix} Thnx In Advance.

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  • $\begingroup$ It's the same as usual: the eignevalues are the roots of the characteristic polynomial $\det(tI - A)$. $\endgroup$ – André 3000 Apr 19 '14 at 20:45
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We find the eigenvalues and eigenvectors as usual.

We have eigenvalues $|A - \lambda I| = 0$ and arrive at:

$$\lambda_{1,2,3} = -2, 0, 0$$

We have eigenvectors from $[A - \lambda_i I]v_i = 0$ and get:

$$v_1 = (-1,1,0), v_2 = (0,0,1), v_3 = (1,1,0)$$

We were lucky and did not need to find generalized eigenvectors.

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  • $\begingroup$ Apart from the third entries can be anything in $\mathbb{R}$ $\endgroup$ – Ellya Apr 19 '14 at 20:53
  • $\begingroup$ Yep, that's all good $\endgroup$ – Ellya Apr 19 '14 at 20:54
  • $\begingroup$ I thought he asked about a general matrix. Not specifically 3x3. $\endgroup$ – Royi Apr 19 '14 at 21:47
  • $\begingroup$ I don't know. That's the question. Maybe it guarantees a specific value of the Eigen Values, maybe specific direction, etc... $\endgroup$ – Royi Apr 19 '14 at 21:52
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Take the characteristic polynomial as usual and we get:

$-\lambda((1+\lambda)^2-1)=0\Rightarrow \lambda(\lambda^2+2\lambda)=0\Rightarrow \lambda=0,0,-2$ are the three eigenvalues.

for $\lambda=0$, we get eigen vectors satisfying: $x=y$, $z=z$, so eigenvectors are of the form $(1,1,\alpha)^T$ for $\alpha\in\mathbb{R}$.

For $\lambda=-2$ we get eigen vectors satisfying: $x=-y$, $z=z$ , so eigenvectors are of the form $(1,-1,\alpha)^T$ for $\alpha\in\mathbb{R}$

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It takes only a very mild form f curiosity to try out what happens if you multiply this matrix with a vector that is nonzero only in its last component.

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