From a Quantum Computing course that I'm following. I am trying to figure out what are the steps to simplify this matrix. The answer is -0.23:
$$ \sqrt{\frac{1}{2}}\begin{bmatrix} \sqrt{\frac{1}{3}}&-\sqrt{\frac{2}{3}}i \end{bmatrix}\begin{bmatrix} 1 & 1 \\ 1 &-1\end{bmatrix}\begin{bmatrix} \sqrt{\frac{1}{3}}\\\sqrt{\frac{2}{3}}i \end{bmatrix} $$
I'm really rusty (bad) with matrix algebra, so I can't find a way to solve this.
The expression is in the form
$$c\cdot \vec v^HA \vec v$$
and the result is a scalar.
To proceed we can at first calculate
$$\vec w=A \vec v=\begin{bmatrix} 1 & 1 \\ 1 &-1\end{bmatrix}\begin{bmatrix} \sqrt{\frac{1}{3}}\\\sqrt{\frac{2}{3}}i \end{bmatrix}=\begin{bmatrix} \sqrt{\frac{1}{3}}+\sqrt{\frac{2}{3}}i \\\sqrt{\frac{1}{3}}-\sqrt{\frac{2}{3}}i \end{bmatrix}$$
then
$$\vec v^HA \vec v=\vec v^H\vec w=\begin{bmatrix} \sqrt{\frac{1}{3}}&-\sqrt{\frac{2}{3}}i \end{bmatrix}\begin{bmatrix} \sqrt{\frac{1}{3}}+\sqrt{\frac{2}{3}}i \\\sqrt{\frac{1}{3}}-\sqrt{\frac{2}{3}}i \end{bmatrix}=\frac13-\frac23=-\frac13$$
and finally
$$c\cdot \vec v^HA \vec v=\sqrt{\frac{1}{2}}\left(-\frac13\right)\approx -0.236$$