I was trying to prove if the Hadamard gate matrix is unitary. Maybe you can help me out in figuring out the conjugate tranpose of this matrix:
$\frac{1}{\surd2} \begin{bmatrix}1 & 1 \\1 & -1 \end{bmatrix}$

My thoughts so far
I suspect that the conjugate transpose is $\frac{1}{\surd2} \begin{bmatrix}-1 & -1 \\-1 & 1 \end{bmatrix}$. If this is right, then multiplying this adjoint to the actual matrix, will result in this matrix $\begin{bmatrix}-1 & 0 \\0 & -1 \end{bmatrix}$. Is this the same as the identity matrix?

  • 1
    $\begingroup$ Your guess is very bizarre. Do you know what the words "conjugate" and "transpose" mean separately in this context? $\endgroup$ – Ben Grossmann Sep 26 '20 at 13:19
  • $\begingroup$ I heard two conflicting methods to go about calculating the conjugate. One is to just flip the signs of the imaginary part of the vector, which would result in the same matrix. The other was from this video: youtube.com/watch?v=oHzpMgKuI9Q&t=117s I used the method from this video $\endgroup$ – Sinestro 38 Sep 26 '20 at 13:28
  • $\begingroup$ Since the matrix is symmetrical I think that my error lies in finding the conjugate instead of the transposing part. $\endgroup$ – Sinestro 38 Sep 26 '20 at 13:29
  • $\begingroup$ That video doesn't seem to have anything to do with complex conjugation. $\endgroup$ – Mark Sep 26 '20 at 13:32
  • $\begingroup$ I thought that this doesn't have to do with complex numbers either. Forgive me, I'm new to linear algebra. I thought that this is a real matrix so that complex conjugation isn't needed. $\endgroup$ – Sinestro 38 Sep 26 '20 at 13:34

This matrix is symmetric and all of its entries are real, so it's equal to its conjugate transpose.

The matrix you are asking about is different from the identity matrix.

But the original matrix is unitary.


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