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Find all complex-valued $n \times n$ matrices $A$ such that $\det A = 0$ and $\text{rank}(AB) = \text{rank}(BA)$ for any $n \times n$ complex-valued $B$.

I believe that $A = 0$ is the only answer.

I have been able to prove that, if $A$ is of rank $r$, then any $r$ lines and any $r$ columns are linearly independent. To see this, note that since $A$ is of rank $r$, then A has $r$ linearly independent columns; say that the indexes of these columns are $i_1, i_2, ..., i_r$. Then by making B equal to a matrix that has 1 in positions $(i_k,i_k)$ and 0 elsewhere, $AB$ basically "selects" $r$ independent columns from $A$ having all other columns equal to 0, so $\text{rank} AB = r$.

Now, $BA$ selects rows $i_1,i_2,..,i_r$ from $A$. If $A$ were to have $r$ rows that were not linearly independent, then there would be an inversible matrix $M$ which would place these rows in positions $i_1,i_2,...,i_r$. Then $$\text{rank} (BA) = \text{rank} (BAM) < r,$$

which would contradict our hypothesis. Thus any $r$ rows of $A$ are linearly independent. Running the same argument in reverse, we get that any $r$ lines of $A$ are linearly independent.

Any ideas about how to proceed?

Tanny Sieben
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  • It gives you perhaps too little information about the matrix. For example, $A=\begin{bmatrix}1 & 1\1 & 1\end{bmatrix}$ is not zero, but it satisfies your conclusion that any row and any column is linear independent ($r=1$). – A.Γ. Apr 14 '19 at 21:02

3 Answers3

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Here is a simple way. Suppose $A\ne 0$ and $\det A=0$. Then its null space $N(A)$ and its range $R(A)$ are both nontrivial. So we can pick any nonzero $v\in N(A)$ and $w\in R(A)$.

Choose any linear map $\mathbf{B}:\mathbb{R}^n\to N(A)\subseteq\mathbb{R}^n$ such that $\mathbf{B}w=v$, and let $B$ be its matrix. Then it is easy to verify that $AB=0$ but $BA\ne 0$.

Eclipse Sun
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  • Wait, does B take just that specific $w$ to that specific $v$? If yes, why do we get that $AB = 0$? Or does $B$ take $\text{range} A$ to $\text{ker} A$? – Tanny Sieben Apr 14 '19 at 21:17
  • I let the range of $B$ be contained in $\ker A$, so $AB=0$. – Eclipse Sun Apr 14 '19 at 21:23
  • Ah, didn't catch that. Please tell me if I got it right: since the range of $B$ is in ker $A$ then $AB = 0$, and since $w$ is in the range of $A$, there is a vector $z$ for which $Az = w$. Then $BAz = Bw = v \neq 0$ so $BA \neq 0$. Only one thing still bugs me: How can we be sure that the linear map $B$ exists? – Tanny Sieben Apr 14 '19 at 21:29
  • Yes. To see the existence, extend $w$ to a basis $w, e_2, \ldots, e_n$ and then you can define $\mathbf{B}$ with the desired property. – Eclipse Sun Apr 14 '19 at 22:02
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Another way to do it: assume $\operatorname{rank}A=r$.

Since both conditions (rank and $\det$) are invariant under pre- and postmultiplication of $A$ by an invertible transformation $\color{red}{(*)}$, it is enough to consider $A$ in the form $$ A=\begin{bmatrix}I_r & 0\\0 & 0\end{bmatrix}. $$ The rank condition becomes $$ \operatorname{rank}\left(\begin{bmatrix}I_r & 0\\0 & 0\end{bmatrix}B\right)= \operatorname{rank}\left(B\begin{bmatrix}I_r & 0\\0 & 0\end{bmatrix}\right). $$ Now it is easy to find a $B$ that contradicts the equality if $0<r<n$.

P.S. Explaining $\color{red}{(*)}$: let $A=L\Sigma R$ where $L$, $R$ are invertible. Then $\det A=0$ iff $\det\Sigma=0$ and $$ \operatorname{rank}\Sigma\, RBL= \operatorname{rank}L\Sigma R\,B= \operatorname{rank}B\,L\Sigma R= \operatorname{rank}RBL\,\Sigma. $$ Take $RBL$ as a new $B$ matrix (arbitrary).

A.Γ.
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  • How can we find such a $B$? If we let $B$ be $\begin{bmatrix}X & Y \ Z & T \end{bmatrix}$, then $AB$ and $BA$ become $\begin{bmatrix} X & Y \ 0 & 0 \end{bmatrix}$ and $\begin{bmatrix} X & 0 \ Z & 0 \end{bmatrix}$ respectively. Now I'm thinking about choosing $X$ to be $I_r$ but with the last row 0; choosing $Y$ to be $0$ but with $1$ somewhere in it's last row; and $Z$ be basically a truncated version of $X$. Then rank $AB$ is $r$ since it has $r$ independent rows, and rank $B$ is $r-1$ since the rows in $Z$ repeat in $X$. Would this work? – Tanny Sieben Apr 14 '19 at 21:42
  • @TannySieben Possibly it is better to choose $X=0$ and $Y$, $Z$ of different ranks, for example, $Y=0$ and $Z\ne 0$. What we have proved is that if $\operatorname{rank}(AB)=\operatorname{rank}(BA)$ for all $B$ then $A$ is either zero matrix or of full rank. – A.Γ. Apr 14 '19 at 21:50
  • Thanks, that sounds much more reasonable. – Tanny Sieben Apr 14 '19 at 21:52
  • @TannySieben You are welcome. – A.Γ. Apr 14 '19 at 21:55
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Here is a reasoning to show just how difficult it is to have the ranks of $AB$ and $BA$ to be equal for all $B$, or even just for all $B$ of rank$~1$ (in making that limitation, my answer is similar to the one by Eclipse Sun). Such rank$~1$ matrices can be obtained as matrix products $B=CR$ where $C,R$ are nonzero column respectively row matrices (i.e., of size $n\times 1$ respectively $1\times n$). Changing $C$ in this expression can always be obtained (in many ways if $n>1$) by left-multiplication by some invertible matrix, and this therefore does not change the rank of $BA$; in the same way changing $R$ can be obtained by right-multiplication by some invertible matrix, and this therefore does not change the rank of $AB$. It follows that if the ranks of $AB$ and $BA$ are to be equal for all such$~B$, both have to be independent of $B$ altogether.

But then then only possibilities are when both ranks are always$~1$, which happens only for invertible matrices $A$ that were excluded by the hypothesis $\det(A)=0$, or when both ranks are always$~0$ which happens only for $A=0$.

In a bit more detail you can see, in the given situation, that the rank of $AB$ equals that of $AC$, and that the rank of $BA$ equals that of $RA$ (both are in $\{0,1\}$), and this quickly gives the indicates reasoning.

Marc van Leeuwen
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