How to find the eigenvalues and eigenvector without computation?

The given matrix is

$$\begin{pmatrix} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2 \\ \end{pmatrix}$$

so, how could i find the eigenvalues and eigenvector without computation? Thank you

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It is still computation, albeit trivial computation. –  André Nicolas Oct 12 '12 at 22:05
Find them the normal way. Once you do so, it should be obvious how to deal with all similar problems. –  Hurkyl Oct 12 '12 at 22:43
You could guess. I mean, you probably will be wrong, but other than computation it may be your only other option. –  Graphth Oct 13 '12 at 1:04

For the eigenvalues, you can look at the matrix and extract some quick informations.

Notice that the matrix has rank one (all columns are the same), hence zero is an eigenvalue with algebraic multiplicity two. For the third eigenvalue, use the fact that the trace of the matrix equals the sum of all its eigenvalues; since $\lambda_1=\lambda_2=0$, you easily get $\lambda_3=6$.

For the eigenvectors corresponding to $\lambda=0$ sometimes it's not hard; in this case it's clear that $v_1=[1,-1,0]$ and $v_2=[0,1,-1]$ (to be normalized) are eigenvectors corresponding to $\lambda=0$. For the eigenvector corresponding to $\lambda_3$, it's not as obvious as before, and you might have to actually write down the system, finding $v_3=[1,1,1]$ (to be normalized).

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Try to solve the following matrix equation, taking into account that your matrix is singular:

$$\begin{pmatrix}2&2&2\\2&2&2\\2&2&2\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}2(x+y+z)\\2(x+y+z)\\2(x+y+z)\end{pmatrix}=2\begin{pmatrix}x+y+z\\x+y+z\\x+y+z\end{pmatrix}=\lambda\begin{pmatrix}x\\y\\z\end{pmatrix}$$

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