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Show that this matrix is unitary and compute its eigenvalues. Unitarily diagonalize this matrix

\begin{bmatrix}0&i&0\\-1&0&0\\0&0&-i\end{bmatrix}

is got the eigenvalues $\lambda = -i, \sqrt{i}, -\sqrt{i}$

but I can't find the eigenvectors

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  • $\begingroup$ What's $\sqrt{i}$? $\endgroup$ – Roland Mar 14 '16 at 6:17
  • $\begingroup$ i think i calculated my determinant wrong $\endgroup$ – Alex Chavez Mar 14 '16 at 6:27
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$$UU^*=\begin{pmatrix}0&i&0\\\!-1&0&0\\0&0&\!-i\end{pmatrix}\begin{pmatrix}0&\!-1&0\\\!-i&0&0\\0&0&i\end{pmatrix}=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}$$

so $\;U\;$ is unitary. Its eigenvalues are

$$\det(xI-U)=\begin{vmatrix}x&\!-i&0\\1&x&0\\0&0&x+i\end{vmatrix}=x^2(x+i)+i(x+i)=(x+i)(x^2+i)=0\iff$$

$$x=-i\,,\,\,\pm\frac1{\sqrt2}\left(1-i\right)\;,\;\;\text{so for example}$$

$$\lambda=-i:\;\;\begin{cases}-i(x+y)=0\\{}\\x-iy=0\end{cases}\;\;\implies\,x=y=0\;,\;\;\text{so for example}\;\;\begin{pmatrix}0\\0\\1\end{pmatrix}$$

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  • $\begingroup$ Traditionally for the determinant I would have evaluated $\begin{vmatrix} -x & -i & 0 \\ 1 & -x & 0 \\ 0 & 0 & i-x \end{vmatrix}$ as my usual equation is $\det(M - xI)$ is the answer the same regardless of the choice or is there a specific reason you went with your particular choice and sign for the determinant $\endgroup$ – frogeyedpeas Mar 14 '16 at 6:26
  • $\begingroup$ @frogeyedpeas Many use that. I'd rather have a polynomial with leading coefficient $\;1\;$ and not $\;-1\;$ , as would happen for odd order matrices, as in this case. $\endgroup$ – DonAntonio Mar 14 '16 at 7:56

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