# Is this matrix equal to the identity matrix?

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?

• What do you think the identity matrix is? – Randall Sep 26 '20 at 14:12
• That’s $\displaystyle\sigma_{y}$: A Pauli Matrix. – Felix Marin Sep 26 '20 at 14:44
• Yes, I'm trying to prove that the Pauli Matrix is unitary. From what I know, a unitary matrix is $\begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix}$ – Sinestro 38 Sep 26 '20 at 18:01

If $$A=\begin{pmatrix}0&-i\\i&0\end{pmatrix}$$, then $$\bar A^T=A$$.

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

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

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 $$A \bar{A^t} = \begin{bmatrix} 0 & -i \\ i &0 \end{bmatrix}^2 = \ \ ?$$

• 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? – Sinestro 38 Sep 26 '20 at 18:00