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Find a basis of complex 3-vectors and complex eigenvalues for the matrix $$ \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix} $$

I found the eigenvalues to be $\lambda = 1, \lambda = -\frac12i\big(\sqrt{3} - i\big), \lambda = \frac12i\big(\sqrt{3}+i\big)$

Not sure how to go about finding the basis vectors though. Perhaps its something simple like the three vectors (columns) $$\begin{bmatrix} 0 & 0 & -i^2 \\ -i^2 & 0 & 0 \\ 0 & -i^2 & 0 \end{bmatrix} $$ But Im really not sure.

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  • $\begingroup$ The eigenvalues are $1,\zeta_3,\zeta_3^2$ with $\zeta_3=e^{2\pi i/3}$. $\endgroup$ Apr 4, 2018 at 14:47

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For each of your $\lambda$'s, find a non-null vector $v$ such that$$\begin{bmatrix}0&0&1\\1&0&0\\0&1&0\end{bmatrix}.v=\lambda v.$$Those $3$ vectors will form the basis that you're after.

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  • $\begingroup$ Isn't that the same as computing the eigenvectors? Is there some intuitive understanding as to why the eigenvectors form a basis for the matrix? $\endgroup$
    – Pame
    Apr 4, 2018 at 14:53
  • $\begingroup$ Yes, that's computing the eigenvectors. I thought that that's what you wanted. It it's not, then please tell me what is a basis of a matrix. $\endgroup$ Apr 4, 2018 at 14:58
  • $\begingroup$ @Pame They don’t form “a basis for the matrix.” They form a basis of $\mathbb C^3$ in which the matrix is diagonal. $\endgroup$
    – amd
    Apr 4, 2018 at 20:15
  • $\begingroup$ Yeah i guess thats what they meant. I just restated excatly what the problem said. I guess 3 linearly independent complex vectors with 3 coordinates will span $\mathbb{C^3}$ $\endgroup$
    – Pame
    Apr 5, 2018 at 21:20

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