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 $BA=2I_2$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.