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I'm trying to prove if this matrix is unitary: $\begin{bmatrix}0 & -i \\i & 0 \end{bmatrix}$
So after multiplying it by it's conjugate transpose I got the answer $\begin{bmatrix}-i & 0 \\0 &-i \end{bmatrix}$
Is this equal to the identity matrix?

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    $\begingroup$ What do you think the identity matrix is? $\endgroup$ – Randall Sep 26 '20 at 14:12
  • $\begingroup$ That’s $\displaystyle\sigma_{y}$: A Pauli Matrix. $\endgroup$ – Felix Marin Sep 26 '20 at 14:44
  • $\begingroup$ Yes, I'm trying to prove that the Pauli Matrix is unitary. From what I know, a unitary matrix is $\begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix}$ $\endgroup$ – Sinestro 38 Sep 26 '20 at 18:01
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If $A=\begin{pmatrix}0&-i\\i&0\end{pmatrix}$, then $\bar A^T=A$.

Now you can check that $A\bar A^T=A^2=I$.

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  • $\begingroup$ So just to clarify, the second matrix I put in the question is equal to the identity matrix right? $\endgroup$ – Sinestro 38 Sep 26 '20 at 18:49
  • $\begingroup$ No it isn't. Because $-i\ne1$. $\endgroup$ – user403337 Sep 26 '20 at 19:01
  • $\begingroup$ Right, thank you. $\endgroup$ – Sinestro 38 Sep 28 '20 at 14:19
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A matrix is unitary if its conjugate transpose is also its inverse.

Call your first matrix $A$. The conjugate transpose of your first matrix is $ \bar{A^t} = \begin{bmatrix} 0 &i \\ -i &0 \end{bmatrix}^T = \begin{bmatrix} 0 & -i \\ i &0 \end{bmatrix}$ So you can see that your matrix is equal to its conjugate transpose. That property is called Hermitian. Have you tried multiplying $ A \bar{A^t} = \begin{bmatrix} 0 & -i \\ i &0 \end{bmatrix}^2 = \ \ ?$

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  • $\begingroup$ I did try that, I actually got the answer I put in my question. Is that right? Also, I heard that another way that you can tell if a matrix is unitary by seeing if the cross product of matrix A and conjugate tranpose A = Identity matrix. Can you verify this? $\endgroup$ – Sinestro 38 Sep 26 '20 at 18:00

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