Is this a metric on matrices? In the set of $n$-by-$n$ reversible real matrices, decide whether 
$$d(A,B)=\ln (\lVert A^{-1}B\rVert\cdot\lVert B^{-1}A\rVert)$$
defines a metric and/or semi-metric.
Can you please help me to solve it?
 A: To check if it is a metric, first you should define what $||A||$ means; it is of course a matrix norm, but which one?
Unfortunately not all norms work; for example, to satisfy the property $d(A, B ) = 0 \iff A = B$ we need $||I|| = 1$ but this is not true for all the matrix norms.
Anyhow you're book seems to assume that your matrix already satisfy the first axioms (that is, $d(A, B) \ge 0$,  $d(A, B) = d(B, A)$ (this is obvious) and $d(A, B) = 0 \iff A = B$) 
Now, you just have to check wether the triangle inequality holds; 
Is $$d(A, B) \le d(A, C) + d(C, B)$$ for every invertible matrix $A, B, C$ ?
Things are greatly simplified if we have a sub multiplicative norm.
Assuming it is, we want to show that 
$$\ln(||A^{-1}B|| \cdot ||B^{-1}A||) \le \ln(||A^{-1}C|| \cdot ||C^{-1}A||) + \ln(||C^{-1}B|| \cdot ||B^{-1}C||)$$
Taking the exponential on both sides
$$||A^{-1}B|| \cdot ||B^{-1}A|| \le ||A^{-1}C|| \cdot ||C^{-1}A|| \cdot ||C^{-1}B|| \cdot ||B^{-1}C||$$
Now, if the norm is multiplicative, write $B = CC^{-1}B$ and get $$||A^{-1}B|| = ||A^{-1}CC^{-1}B|| \le ||A^{-1}C||\cdot ||C^{-1}B||$$
Now the same trick, with $A = CC^{-1}A$
$$||B^{-1}A|| = ||B^{-1}CC^{-1}A|| \le ||B^{-1}C|| \cdot ||C^{-1}A||$$
and substituting you get indeed $1 \le 1$ which is true.
To recap:
1) It is not a distance for every norm. You should at least have, for example $||I|| = 1$
2) Anyhow your book seems to assume that the first axioms are respected, so you just need to check wether it respect the triangle inequality or not. For a general norm I have not been able to find an answer, but for a sub multiplicative norm then you get that $d(A, B)$ is indeed a metric because it respect the triangle inequality (as shown above)
