Invertibility of product of two matrices

Let $A$ be a $m \times n$ matrix and $B$ be a $n \times m$ matrix over real number with $m < n$.

Which one is correct ?

1. $AB$ is always singular

2. $BA$ is always singular

Option $1$ is true because take $$A=\begin{pmatrix} 1&&0&&0\\0&&1&&0\end{pmatrix}$$ and $$B=\begin{pmatrix} 1&&0\\0&&1\\0&&0\end{pmatrix}$$ Then $AB = I$

Option 2 is not True

• Similar question -math.stackexchange.com/questions/2838189/… – BAYMAX Jul 14 '18 at 14:37
• From your question it seems that you don't use the definition of a singular matrix in a correct way...is the unity matrix invertible? – user190080 Jul 15 '18 at 13:26

$BA$ is always singular!

Hint:

$A$ is $m \times n$ and $m<n$ implies $Ax=0$ has a non-zero solution

$\textbf{Added}$:

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.

• im not getting can u elaborate more,,,,,@Chinnapparaj – Messi fifa Jul 14 '18 at 14:03
• Let $x_0$ be the non zero solution of $Ax=0$. The same $x_0$ will be the solution to $(BA)x=0$. Can you see why? – Chinnapparaj R Jul 14 '18 at 14:05
• I edit my answer and now i think you are ok! – Chinnapparaj R Jul 14 '18 at 14:13
• For your first part, we do not conclude anything about $AB$. It may be singular or non singular. For your example, it is non singular. On the other hand consider $A=\begin{pmatrix} 1& -1 \end{pmatrix}$ and $B=\begin{pmatrix} -1\\ -1 \end{pmatrix}$. Then $AB=0$ – Chinnapparaj R Jul 14 '18 at 14:18
• You are welcome! – Chinnapparaj R Jul 14 '18 at 14:22