I wanted to know if I did this problem right or not.
$A$ and $B$ are the following matrices:
$A=\begin{bmatrix}2&6\\1&-3\end{bmatrix},$
$B=\begin{bmatrix}3&2\\1&0\end{bmatrix}.$
Then
$$\frac{\langle A,B\rangle}{\|A\|\|B\|} =\frac{\left\langle\begin{bmatrix}2&6\\1&-3\end{bmatrix}, \begin{bmatrix}3&2\\1&0\end{bmatrix}\right\rangle}{\sqrt{\left\langle\begin{bmatrix}2&6\\1&-3\end{bmatrix},\begin{bmatrix}2&6\\1&-3\end{bmatrix}\right\rangle}\sqrt{\left\langle\begin{bmatrix}3&2\\1&0\end{bmatrix},\begin{bmatrix}3&2\\1&0\end{bmatrix}\right\rangle}}.$$
I ended up getting:
$\arccos(19/184) = 84.07^\circ$
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$\begingroup$$a\cdot b=|a||b|\cos\theta$ so
$$\theta=\arccos\Big(\frac{a\cdot b}{|a||b|}\Big).$$
we have
$a\cdot b=6+12+1=19, |a||b|=\sqrt{4+36+1+9}\sqrt{9+4+1}=10\sqrt{7}$.
so the angle is $\arccos(19/(10\sqrt{7}))$
I realize this is old, but there's a mistake in the answer. The process is generally correct, except that the angle calculated depends on the rank of the tensors. For a 1st rank tensor, like a vector, the above product is equal to the first Legendre polynomial, cos(theta). For a second rank, it is equal to the second Legendre polynomial, 0.5(3*cos(theta)^2 -1), and so on.
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