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

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

• What's $\sqrt{i}$? Mar 14 '16 at 6:17
• i think i calculated my determinant wrong Mar 14 '16 at 6:27

$$UU^*=\begin{pmatrix}0&i&0\\\!-1&0&0\\0&0&\!-i\end{pmatrix}\begin{pmatrix}0&\!-1&0\\\!-i&0&0\\0&0&i\end{pmatrix}=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}$$
so $\;U\;$ is unitary. Its eigenvalues are
$$\det(xI-U)=\begin{vmatrix}x&\!-i&0\\1&x&0\\0&0&x+i\end{vmatrix}=x^2(x+i)+i(x+i)=(x+i)(x^2+i)=0\iff$$
$$x=-i\,,\,\,\pm\frac1{\sqrt2}\left(1-i\right)\;,\;\;\text{so for example}$$
$$\lambda=-i:\;\;\begin{cases}-i(x+y)=0\\{}\\x-iy=0\end{cases}\;\;\implies\,x=y=0\;,\;\;\text{so for example}\;\;\begin{pmatrix}0\\0\\1\end{pmatrix}$$
• 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 Mar 14 '16 at 6:26
• @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. Mar 14 '16 at 7:56