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M =$\left[\begin{array}{ccc}1& 1 & 1 \\a & b & c \\d & e & f\end{array}\right]$.

Work so far:

Let $\lambda$ denote the eigenvalues. By my calculations, the eigenvalues must $\lambda = 3,0,0$.

$M_1 =\left[\begin{array}{ccc}-2 & 1 & 1 \\a & b-3 & c \\d & e & f-3\end{array}\right]$.$M_1*x = (3,3,3)^T$.

$M_2 =\left[\begin{array}{ccc}1& 1 & 1 \\a & b & c \\d & e & f\end{array}\right]$. $M_2x = (0,0,0)^T$.

I'm not sure where to go from here.

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  • $\begingroup$ Since $M$ is rank $1$, each row is a multiple of the other $\endgroup$ – Omnomnomnom Apr 9 '15 at 19:38
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Note that since $\left [ \begin{array}{ccc} 1&1&1 \\ \end{array} \right ]^T$ is an eigenvector of the above matrix you must have that

$$ \left [ \begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ d & e & f \\ \end{array} \right ] \left[ \begin{array}{c} 1 \\ 1 \\ 1 \\ \end{array} \right ] \;\; =\;\; \left [ \begin{array}{c} 3 \\ a + b + c \\ d + e + f \\ \end{array} \right ] \;\; =\;\; \lambda\left [ \begin{array}{c} 1\\ 1\\ 1\\ \end{array} \right ] $$

hence $\lambda = 3 = a+b+ c = d +e +f$ is one eigenvalue. Applying the same idea to the second vector $\left [ \begin{array}{ccc} 1 & 0 & -1 \\ \end{array} \right ]^T$ we obtain

$$ \left [ \begin{array}{c} 0 \\ a - c \\ d - f \\ \end{array} \right ] \;\; =\;\; \lambda\left [ \begin{array}{c} 1 \\ 0 \\ -1 \\ \end{array} \right ] $$

so $\lambda = 0$ is another eigenvalue constraining $a = c$ and $d = f$. Can you follow the pattern now?

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  • $\begingroup$ @larry I edited my answer to reflect the computation. The second line follows the exact same computation. $\endgroup$ – Mnifldz Apr 9 '15 at 19:39
  • $\begingroup$ @larry You have one more eigenvalue/eigenvector pair to compute. Try mimicking what I did above to reveal any more constraints on $a,b,c$ and $d,e,f$. You need three equations to solve three unknowns. $\endgroup$ – Mnifldz Apr 9 '15 at 19:44
  • $\begingroup$ got it. thanks. Just forgot about the last eigenvalue/eigenvector.. $\endgroup$ – larry Apr 9 '15 at 19:50

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