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For the inner product space $M_{2\times2} (\Bbb{C})$, with inner product $\langle A,B\rangle = \operatorname{tr}(AB^*)$, find the angle between

$$A=\begin{bmatrix}2+i&1\\-1&0\end{bmatrix} $$

$$B=\begin{bmatrix}3-2i&1+i\\0&-1\end{bmatrix}$$

I tried to use $$\cos(z)=\frac{\langle A,B\rangle}{\|A\|\|B\|}=\frac{5+6i}{4\sqrt{7}}$$ but how can I find the angle? is it supposed to be a real number?

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    For that matter, with the standard (complex) inner product on ${\bf C}^2$, how do you find the angle between two vectors? – 2017-01-22
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    See my answer [here](http://math.stackexchange.com/questions/2079945/what-is-the-angle-between-those-two-matrices-over-mathbbc/2080001#2080001) – 2017-01-22

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The inverse of the cos function is defined for complex arguments as well.

$$\operatorname{arccos}(z)=-i\operatorname{Log}\left(iz+\sqrt{1-z^2}\right).$$

Where Log is the so called complex logarithm defined as

$$\operatorname{Log}(u)=\ln(\mid u\mid)+i\operatorname{atan2}(\Im(u),\Re(u)).$$

And finally $\operatorname{atan2}$ is defined as

enter image description here

So the complex angle in the OP's case would be a complex number:

$$\operatorname{arccos}\left(\frac{5+6i}{4\sqrt{7}}\right).$$

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    OK, but is that really the way to define the angle between two vectors in a complex inner product space? – 2017-01-23
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    To be honest, I don't know. But I saw that something analogue to the angle could be calculated. I cannot intuit this angle between conplex matrice5s. I just hoped that you knew why you tried to compute such things. – 2017-01-23