Is the Transpose Conjugate of a Unitary Matrix Equal to the Identity Matrix?

[andphy]

Hi,

A unitary matrix should have it transpose conjugate equal to its inverse. Please confirm that this statement is correct and check attached matrix as they are not equal and in doubt if I did correctly.

Thanks.

 

[Clear Mind]

Hi,
suppose to have an unitary matrix ($U \in \mathbb{C}^{n \times n}$, so that $U^{\dagger}=U^{-1}$), if you want to verify the unitariety of your matrix, just check if $UU^{\dagger}= \mathbb{I}$.

 

[andphy]

That helps thank you.

 

[Fredrik]

Are you saying (in the attached file) that $\begin{pmatrix}i & 0\\ 0 & 1\end{pmatrix}$ is the inverse of $\begin{pmatrix}1 & 0\\ 0 & i\end{pmatrix}$? It's not. To see this, just multiply these two matrices together.

 

[HallsofIvy]

You have at one point, that the "transpose conjugate" of \begin{bmatrix}1 & 0 \\ 0 & i\end{bmatrix} is \begin{bmatrix}1 & 0 \\ 0 & -i\end{bmatrix}. That is correct and that is the inverse matrix.

Below that, you have "inverse matrix" and \begin{bmatrix} i & 0 \\ 0 & 1\end{bmatrix}. I don't know where that came from!

 

[andphy]

Right the product will result in the inverse:

1 0
0 1

correct ?

 

[Fredrik]

Right the product will result in the inverse:

1 0
0 1

correct ?

What product? The product of the two matrices in post #5 is $\begin{pmatrix}i & 0 \\ 0 & i\end{pmatrix}$. As HallsofIvy said, the inverse of $\begin{pmatrix}1 & 0\\ 0 & i\end{pmatrix}$ is $\begin{pmatrix}1 & 0\\ 0 & -i\end{pmatrix}$. The product of these two matrices is the identity matrix.

 

[andphy]

sorry meant to say identity matrix (not inverse) - thank you.