Suppose $A, B$ respectively be a matrix with size $4\times 2$ and $2\times4$. If $AB = \begin{pmatrix} 1&0&-1&0\\0&1&0&-1\\-1&0&1&0\\0&-1&0&1\end{pmatrix} \\$ then $BA$ is equal to what?
For this one, I really don't have much idea to think. Maybe eigen things work here because sometime it shows miracles. Could you help me? I have discussed this one week ago with my friend in college, but didn't find the way
The columns of $A$ must be able to combine to create the row space of $AB$, and this row space is easy to imagine due to the many zeroes in $AB$: We must have
$$ A = \begin{pmatrix}1 & 0 \\ 0 & 1 \\ -1 & 0 \\ 0 & -1 \end{pmatrix} P $$
where $P$ is some invertible $2\times 2$ matrix. Then, however, we can compute how $B$ must look to get the right product, namely
$$ B = P^{-1} \begin{pmatrix} 1 & 0 & -1 & 0 \\ 0 & 1 & 0 & -1 \end{pmatrix} $$
Putting those together we get
$$ BA = P^{-1} \begin{pmatrix} 1 & 0 & -1 & 0 \\ 0 & 1 & 0 & -1 \end{pmatrix} \begin{pmatrix}1 & 0 \\ 0 & 1 \\ -1 & 0 \\ 0 & -1 \end{pmatrix} P $$
which you can multiply out and see that the $P$s cancel out.