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Let $A$ be a $4\times2$ matrix and $B$ be a $2\times4$ matrix so that $$AB= \begin{pmatrix} 1 & 0 & -1 & 0 \\ 0 & 1 & 0 & -1 \\ -1 & 0 & 1 & 0 \\ 0 & -1 & 0 & 1 \end{pmatrix}. $$ Find $BA$.
I have the answer, which should be $BA= \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} $, but how do I show that this is the only possible solution or is it sufficient (according to some property which I'm not aware of) to have one match for $A$ and $B$ and therefore no other outcome for $BA$ is possible.
Write $A=\begin{pmatrix}A_1\\A_2 \end{pmatrix}$ and $B=\begin{pmatrix}B_1 &B_2 \end{pmatrix}$ where $A_1,A_2,B_1,B_2$ are $2\times2$ matrices.
Note that $AB=\begin{pmatrix}A_1B_1 & A_1B_2\\A_2B_1 & A_2B_2 \end{pmatrix}$, hence $A_1B_1= A_2B_2=I_2$.
Since a square matrix commutes with its inverse, we have as well $B_1A_1=B_2A_2=I_2$.
Finally, note that $BA=\begin{pmatrix}B_1A_1+B_2A_2 \end{pmatrix}=\begin{pmatrix}2I_2 \end{pmatrix}=\begin{pmatrix}2 & 0 \\ 0 & 2\end{pmatrix}$.
You may compute the square of $AB$ and note that $ AB\; AB = 2 AB$ which means that $A(BA -2I)B=0$. Now the rank of $AB$ is two so both $A$ and $B$ must have rank $2$. Therefore $BA-2I$ is the zero matrix.
Let $C = (c_{ij}) = AB$, so $c_{ij} = a_{i1}b_{1j}+a_{i2}b_{2j}$, that means each entry $c_{ij}$ can be interpreted as the scalar product of row $i$ of $A$ and column $j$ of $B$. Let $a_{i:}$ be the $i$th row of $A$ and $b_{:j}$ the $j$-th column of $B$. Now we see that
$$a_{1:} \perp b_{:2} \text{ and } a_{1:} \perp b_{:4}$$
$$a_{3:} \perp b_{:2} \text{ and } a_{3:} \perp b_{:4}$$
$$a_{2:} \perp b_{:1} \text{ and } a_{4:} \perp b_{:1}$$
$$a_{2:} \perp b_{:3} \text{ and } a_{4:} \perp b_{:3}$$
and therefore $a_{1:} \perp b_{:2} \perp a_{3:}$ which implies $a_{1:} \Vert a_{3:}$ and similarly $a_{2:} \Vert a_{4:}$ and $b_{:1} \Vert b_{:3}$ and $b_{:2} \Vert b_{:4}$.
So we can conlcude $a_{1:} = c a_{3:}$ for some factor $c$. But we know that $a_{1:} \cdot b_{:1} = 1$ and $ca_{1:} \cdot b_{:1} = a_{3:} \cdot b_{:1} = -1$ so $c=-1$.
In the very same way we can conclude that $a_{2:} = -a_{4:}$ and $b_{:1} = - b_{:3}$ and $b_{:2} = - b_{:4}$. To summarize what we've got so far:
$$A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ -a_{11} & -a_{12} \\ -a_{21} & -a_{22} \end{bmatrix} \quad B = \begin{bmatrix} b_{11} & b_{12} & -b_{11} & -b_{12} \\ b_{21} & b_{22} & -b_{21} & -b_{22} \end{bmatrix}$$
Now consider the blocks $$\tilde A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \text{ and }\tilde B = \begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{bmatrix} .$$
Note that $A = \begin{bmatrix} \tilde A \\ - \tilde A \end{bmatrix}$ and $B = \begin{bmatrix} \tilde B & -\tilde B\end{bmatrix}$.
You immediately see from the given equation that $\tilde A \tilde B = I$ so $\tilde A = \tilde B^{-1}$.
And therefore $BA = \tilde B \tilde A + (-\tilde B)(-\tilde A) = 2I$ which is your desired result.
The characteristic polynomial of $AB$ equals $X^2(X-2)^2$. But we have $X^2P_{BA}(X)=P_{AB}(X)$ (see here), and then $P_{BA}(X)=(X-2)^2$.
Now assuming that the given matrices are over a field of characteristic $\ne 2$ we can conclude that $BA$ is invertible.
Moreover, by multiplying $(AB)^2=2(AB)$ with $B$ on the left and with $A$ on the right we get $(BA)^3=2(BA)^2$, and therefore $BA=2I_2$.
