I have $A\in \mathbb{R}^{3\times 2}, \ B\in \mathbb{R}^{2\times 3}\;$ such that $AB=\begin{bmatrix} 8 & 2 & -2\\ 2 & 5 & 4\\ -2 & 4 & 5 \end{bmatrix}$
How can I show that $BA=\begin{bmatrix} 9 & 0\\ 0 & 9 \end{bmatrix}$?
I tried by bruteforcing systems of linear equations, but it seems too complex.
Note that $AB$ is diagonalizable with eigenvalues $0,9$. For the eigenvalue $\lambda = 9$, the associated eigenspace $E_9(AB)$ is two-dimensional. Note that for any vector $x$ from this eigenspace, we have $$ (BA)(Bx) = B(AB)x = B\cdot 9x = 9(Bx). $$ That is: for any $x \in E_9(AB)$, $Bx \in E_9(BA)$.
The remaining eigenvalue of $AB$ is $0$. However, we note that $A$ must have full rank (why?) and therefore has a trivial kernel. Thus, if $x \in E_0(AB)$ so that $ABx = A(Bx) = 0$, then it follows that $Bx = 0$.
Putting these facts together, we see that for $x$ taken from either $E_9(AB)$ or $E_0(AB)$, it holds that $BA(Bx) = 9(Bx)$. Thus, it holds for any $x$ that $BA(Bx) = 9(Bx)$. That is, for any $y$ taken from the range (AKA image/column space) of $B$, $BAy = 9y$. However, $B$ must have full rank (why?), which means the range of $B$ must be all of $\Bbb R^2$. Thus, $BAy = 9y$ for all $y \in \Bbb R^2$.
It follows that $BA = 9I$, as desired.
Alternative proof: we observe as above that $AB$ is diagonalizable. In particular, there exists an invertible matrix $S$ such that $$ S^{-1}(AB)S = \pmatrix{9&0&0\\0&9&0\\0&0&0} = \pmatrix{9I_2 & 0\\ 0 & 0}. $$ Let $D$ refer to the above diagonal matrix. Let $P = S^{-1}A$, and let $Q = BS$. We see above that $PQ = D$, and to prove the desired result it suffices to show that $QP = 9I$.
We now partition $Q,P$ for block matrix multiplication. In particular, we have $$ PQ = D \implies \pmatrix{P_1\\P_2} \pmatrix{Q_1 & Q_2} = \pmatrix{9I_2 & 0 \\0 & 0}. $$ In the above, both $P_1$ and $Q_1$ are $2\times 2$. If we compute the block-matrix product on the left, we find that $$ \pmatrix{P_1Q_1 & P_1 Q_2\\ P_2 Q_1 & P_2Q_2} = \pmatrix{9I_2 & 0 \\0 & 0}. $$ If we consider the top-left entries, we see that $P_1Q_1 = 9I \implies Q_1 = 9 P_1^{-1}$. Looking at the top-right entry, we have $P_1Q_2 = 0$. Because $P_1$ is invertible, we conclude that $Q_2 = 0$. Similarly, we conclude from $P_2Q_1 = 0$ that $P_2 = 0$.
With that, we find that $$ QP = \pmatrix{Q_1\\ Q_2} \pmatrix{P_1 & P_2} = Q_1P_1 + Q_2 P_2 = (9P_1^{-1})P_1 + 0\cdot 0 = 9I, $$ which was the desried conclusion.
Note: It's interesting to compare these two proofs. For instance, the observation $Q_2 = 0$ corresponds to the observation that the image of $B$ lies within $E_{9}(AB)$, and the observation that $P_2 = 0$ corresponds to the observation that the image of $A$ is equal to that of $AB$. The relationship between $P_1,Q_1$ tells us about the behavior of $AB$ over $E_9(AB)$.
Characteristic polynomial of $AB$ is $p_{AB}(x) = x(x-9)^2$.
Characteristic polynomials of $AB$ and $BA$ have this relation: $$p_{AB}(x) = p_{BA}(x)x$$ as explained here. Therefore $$p_{BA}(x) = (x-9)^2.$$
On the other hand, the minimal polynomial of $AB$ is $m_{AB}(x) = x(x-9)$ so $$AB(AB - 9I) = 0 \implies ABAB = 9AB \implies BABABA = 9BABA$$ so te polynomial $x^3-9x^2 = x^2(x-9)$ annihilates $BA$. We know that the minimal polynomial of $BA$ is of the form $m_{BA}(x) = (x-9)^k$ for $k=1, 2$ since $9$ is the only eigenvalue of $BA$. It also must divide $x^2(x-9)$ so it follows that $$m_{BA}(x) = x-9$$ and hence $BA = 9I$.
