An exercise in a textbook I'm using to brush up on my linear algebra asks to prove that the inverse of a persymmetric matrix is also persymmetric. I have a colleague's old notes in front of me with a solution, but I can't understand his reasoning. It states:
Let $A_S$ denote the persymmetric matrix. Then $A = SA_S = A_SS$, where $A$ is the anti-diagonal identity matrix (i.e.: the identity matrix rotated 90°).
$$(SA_S)^{-1} = A_S^{-1}S^{-1}$$
In a previous question, we have proven that $A = A^{-1}$ ($A^2 = I$, where $I$ is the identity matrix, and thus $A^2 = AA = I$, and so multiplying both sides by the inverse $A^{-1}$ we have $A = A^{-1}$). Thus:
$$(SA_S)^{-1} = A_S^{-1}S^{-1} = A_S^{-1}S$$
Therefore we need only prove $A_S^{-1}$ is symmetric. In a previous question we proved the transpose of the inverse of a matrix is equivalent to the inverse of its transpose, so $(A_S^{-1})^T = (A_S^T)^{-1}$. Since we know $A_S^T = A_S$, $(A_S^T)^{-1} = A_S^{-1}$, therefore $A_S^{-1}$ is symmetric. QED.
I can't follow several bits of reasoning in this answer.
First of all, I don't see how the second equation follows from the setup--how does $A = A^{-1}$ tell us that $A_S^{-1}S^{-1} = A_S^{-1}S$?
Why is proving $A_S^{-1}$ to be symmetric sufficient to prove $A_S^{-1}$ is persymmetric?
The author claims $A_S^T = A_S$, which would be true for a symmetric matrix, but not a persymmetric one (it is true that the transpose of a persymmetric matrix is also persymmetric). What is the basis for this claim?
It's also possible that my colleague's answer is incorrect. I have an idea that perhaps using the facts that $A$ rotated by 90° is $I$ and $A_S$ rotated by 90° is a symmetric matrix in order to tackle the proof, but I'm stuck on the specifics.
Can someone either explain my colleague's answer, give me a hint on how to prove this statement, or provide a proof of their own?