<s>[INST] $A^2 - B^2$ is an invertible matrix and $A^5 = B^5$ and $A^3B^2 = A^2B^3$ . Then what is the determinant of the matrix $A^3 + B^3$? $A,B$ are two $3 \times 3$ real-matrices with following three properties.

*

*$A^2 - B^2$ is an invertible matrix

*$A^5 = B^5$

*$A^3B^2 = A^2B^3$
Then what is the determinant of the matrix $A^3 + B^3$ ?
 [/INST] Let $V=\mathbb R^3$. The condition $A^5=B^5$ implies that $A^5V=B^5V$. Since $A$ is $3\times 3$, the descending chain $V\supseteq AV\supseteq A^2V\supseteq\cdots$ must have been stabilised at $A^3V$ and the similar is true for $B^3V$. Therefore from $A^5V=B^5V$ we get $W:=A^3V=B^3V$.
Furthermore, as the descending chains have been stabilised, both $A$ and $B$ are automorphisms on the invariant subspace $W$. The condition $A^3B^2=A^2B^3$ thus implies that $A=B$ on $W$. As $A^2-B^2$ is invertible, $W$ must be zero. Hence $A^3V=B^3V=0$, i.e., $A^3=B^3=0$. Consequently $\det(A^3+B^3)=0$.
Remark. Actually, for any number $c$, “$\det(A^3+B^3)=c$” is a correct answer, because no matrices $A$ and $B$ can satisfy all three given conditions in the first place. Suppose the contrary. By our previous argument, $A$ and $B$ must be nilpotent. Since the matrices are $3\times3$, the ranks of $A^2$ and $B^2$ are at most one. Therefore the rank of $A^2-B^2$ is at most two. Hence $A^2-B^2$ is singular, but this is a violation of the first condition.
However, if we change the first condition to “$A-B$ is invertible”, there are feasible pairs of $A$ and $B$, such as
$$
A=\pmatrix{0&2&0\\ 0&0&1\\ 0&0&0}
\text{ and }B=\pmatrix{0&1&0\\ 0&0&0\\ 1&0&0}.
$$
In this case, our previous argument still applies and we may conclude that $A^3=B^3=0$ and $\det(A^3+B^3)=0$.
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