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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.

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closed as off-topic by José Carlos Santos, copper.hat, Jendrik Stelzner, Adrian Keister, Leucippus Sep 13 '18 at 0:48

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  • $\begingroup$ Just take it one step at a time. Remember that these operations are associative... that is to say $a\cdot b \cdot c\cdot d = ((a\cdot b)\cdot c)\cdot d=a\cdot (b\cdot (c\cdot d)))$. If you are unable to do even one of the steps, then that is where an explanation should start. Ignoring the somewhat messy numbers for now, do you know how to do the multiplication $\begin{bmatrix}a&b\\c&d\end{bmatrix}\cdot \begin{bmatrix}e\\f\end{bmatrix}$? $\endgroup$ – JMoravitz Sep 12 '18 at 23:06
  • $\begingroup$ This just follows the usual rules for matrix multiplication, see en.wikipedia.org/wiki/Matrix_multiplication#Definition for example. You need to do little work. $\endgroup$ – copper.hat Sep 12 '18 at 23:07
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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

$$c\cdot \vec v^HA \vec v=\sqrt{\frac{1}{2}}\left(-\frac13\right)\approx -0.236$$

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