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Let $A$ be an $m \times n$ matrix and $B$ be an $n \times m$ matrix where $m<n$.

Then can we say that the product $AB_{m \times m}$ is always singular or always non-singular?

Also, can we say that $BA_{n \times n}$ is always singular or non-singular?. Does this change any thing?

I was thinking that since $m<n$ we have Rank$(A) \leq m$ and similarly Rank$(B) \leq n$ and also that Rank$(AB) \leq min($Rank$(A)$,Rank$(B)$), but will that help?

How can I think about this problem?

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    $\begingroup$ It may or may not be singular. $\endgroup$ – Lord Shark the Unknown Jul 2 '18 at 6:13
  • $\begingroup$ Consider the special case $A=\begin{bmatrix}M&0\end{bmatrix}$ and $B=\begin{bmatrix}N\\0\end{bmatrix}$, where $M$ and $N$ are $m\times m$ matrices and the zero matrices are of appropriate sizes. What is $AB$ in this case? Can you think of examples where it is singular or nonsingular? $\endgroup$ – Rahul Jul 2 '18 at 6:17
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We have

  • $\text{rank}(A)\le m$, since $A$ has $m$ rows.$\\[4pt]$
  • $\text{rank}(B)\le m$, since $B$ has $m$ columns.

so $$\text{rank}(BA)\le\min(\text{rank}(A),\text{rank}(B))\le m < n$$ hence, since $BA$ is an $n{\,\times\,}n$ matrix, $BA$ is singular.

On the other hand, $AB$ may or may not be singular.

All we need is one example of each . . .

To keep it simple, let $m=1,\,n=2$.

For an example where $AB$ ends up being singular, let $A,B$ be given by $$ A= \pmatrix { 1 & 0\cr } ,\;\;\; B= \pmatrix { 0\cr 1\cr } $$ so we have $\text{rank}(A)=\text{rank}(B)=1$, but $AB=0$, which is singular.

For an example where $AB$ ends up being non-singular, let $A,B$ be given by $$ A= \pmatrix { 1 & 0\cr } ,\;\;\; B= \pmatrix { 1\cr 0\cr } $$ so we have $\text{rank}(A)=\text{rank}(B)=1$, and $AB=(1)$, which, as a $1{\,\times\,}1$ matrix, is non-singular.

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  • $\begingroup$ If $A_{m \times n}$ matrix then can I say that Rank(A) $\leq $ min($m,n$) ? $\endgroup$ – BAYMAX Jul 2 '18 at 13:36
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    $\begingroup$ @BAYMAX: Yes, since the the number of linearly independent rows is equal to the number of linearly independent columns. $\endgroup$ – quasi Jul 2 '18 at 13:48
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$\textbf{BA is always Singular}$:

Result: $A$ is $m \times n$ and $m<n$ implies $Ax=0$ has non zero solution $x_0$.

Proof:

View the matrix $A$ as a linear map $A:\Bbb{F}^n \rightarrow \Bbb{F}^m, x \mapsto Ax.$

By dimension theorem, $$dim\;\Bbb{F}^n=rank\;A+null\;A $$ $$n\leq m+null\;A$$ So, $null\;A \geq n-m >0$,. Hence $Ax=0$ has a non zero solution (say $x_0$). QED

Now $$(BA)x_0=B(Ax_0)=B.0=0$$ Hence $BAx=0$ has non zero solution, concluding $BA$ is singular.

$\textbf{For AB, it may or may not (as in the comment):}$

For example for $m=2$ and $n=3$, Consider

$A=\begin{pmatrix} 1& 0 & 0\\ 0 &1 &0\end{pmatrix}$ and $B=\begin{pmatrix} 1& 0 \\ 0 &1\\ 2 & 2\end{pmatrix}$. Then $AB=I_2$

For the other one, consider $A=\begin{pmatrix} 1& 2 \end{pmatrix}$ and $B=\begin{pmatrix} 2\\ -1 \end{pmatrix}$. Then $AB=0$

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  • $\begingroup$ Nice solution!! $\endgroup$ – BAYMAX Jul 2 '18 at 15:51

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