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Let $m,n\in\Bbb N$ and let $A$ be a matrix such that $$A^mBA^n=I$$ where $$B=\begin{pmatrix}1&-1&3&1\\ 1&1&2&1\\ 2&-1&3&2\\ -1&-2&1&2 \end{pmatrix}$$

$1)$ Is $A$ invertible?

$2)$ Determine $A^{m+n}$.

I'm completely lost...Okay, I know that I need to see whether $\det A=0$ for the first question, and for the second one I know $A^{m+n}=A^m\cdot A^n$. But how do I even start here? I would really appreciate some hint, thanks!

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    What do you know about the determinant of a product of matrices? – 2017-01-31
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    @Student You mean $\det (A^mBA^n)=1 \Rightarrow \det A^m\cdot \det B\cdot \det A^n=1$ therefore they are all $\neq 0$? – 2017-01-31
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    You could still write out the determinant of $A^n$ in terms of the determinant of A but what you wrote is also correct :) – 2017-01-31
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    @Dietrich interestingly $n$ or $m$ being zero is the only case where $B$ and $A$ can fail to be invertible (only works in infinite dimensional setting), if you consider for example left and right shift on $\ell^2(\Bbb N)$ you have $LR=\Bbb 1$ but neither is invertible. – 2017-01-31

2 Answers 2

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$A$ is invertible for a bunch of reasons, for example because the rank of a product of matrices is bounded by the rank of the factors, so $4\geq \text{rank}(A)\geq 4$. Another way to see it is that the determinant of a product of matrices is the determinant of the products, we conclude $\det(A)\neq 0$ and so $A$ is invertible.

Using this:

$A^mBA^n=I\implies A^mB=A^{-n}\implies B=A^{-m}A^{-n} =A^{-m-n}\implies A^{m+n}=B^{-1}$

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    Presupposes $A$ is invertible, but this follows from $A^n$ necessarily being injective and $A^m$ necessarily being surjective. – 2017-01-31
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    Perhaps you should also point out that this can only be done if A is indeed invertible – 2017-01-31
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    Note that there are a bunch of degenerate cases that we're not given in this problem statement that you can mention if you want to be a teachers' pet. For example if $m = n = 0$ then we don't know anything about $A$ and any claim about $A^0 = I$ seems a little sketchy in the off-chance that $A$ is the zero matrix; if $(m, n) \in \{(0, 1), (1, 0)\}$ then we do not necessarily know from the problem description that $A$ is a square matrix hence it might not have a determinant (and might only have, say, a left-inverse and not a right-inverse). – 2017-01-31
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1)It is, and its inverse is $A^m\dot{}B\dot{}A^{n-1}$

2)well explained by Jorge Fernández

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    nice, although how can you show that the inverse is bilateral? – 2017-01-31
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    @ Jorge Fernandez Hidalgo: Using the same trick on the left side we have that A has a right and left inverse – 2017-01-31
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    @Jorge Fernández Hidalgo you can prove that a lateral inverse is the bilateral, right?, I mean in any ring, so you just need to know that $A$ is square, what is implicit in the expression $A^m$ – 2017-01-31
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    not in any ring, although it works for any finite dimensional K-algebra, and also for normal operators – 2017-01-31
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    right, that's a property of groups; my mistake – 2017-01-31