If $A$ is a $2\times 1$ matrix and $B$ is a $1\times 2$ matrix, show that $AB$ is singular 
If $A$ is a $2\times 1$ matrix and $B$ is a $1\times 2$ matrix, show that $AB$ is singular

Here singular means there is no inverse.
I faced this question in my exam and I didn't know how to solve but I have known from my teacher that the multiply $AB$ will create $2\times 2$ matrix and it's easy to prove that it's singular. But also I didn't know how start steps to prove that, I really need help to understand this question because maybe my teacher can repeat it again in another exam.
Thanks in advance.
 A: This clearly not true. Assuming the matrices are over the real numbers
(or any field), note that
\begin{align*}
AB & =\left[\begin{array}{cc}
a_{1} & a_{2}\end{array}\right]\left[\begin{array}{c}
b_{1}\\
b_{2}
\end{array}\right]\\
 & =a_{1}b_{1}+a_{2}b_{2}.
\end{align*}
This element has an inverse unless $a_{1}b_{1}+a_{2}b_{2}=0$.
On
the other hand,
\begin{align*}
\text{det}\left(BA\right) & =\det\left(\left[\begin{array}{c}
b_{1}\\
b_{2}
\end{array}\right]\left[\begin{array}{cc}
a_{1} & a_{2}\end{array}\right]\right)\\
 & =\det\left[\begin{array}{cc}
a_{1}b_{1} & a_{2}b_{1}\\
a_{1}b_{2} & a_{2}b_{2}
\end{array}\right]\\
 & =a_{1}b_{1}a_{2}b_{2}-a_{1}b_{2}a_{2}b_{1}\\
 & =0.
\end{align*}
So $BA$ is singular.
A: Notice that  :


*

*$Rank(A)\leq min\{2,1\}=1$ 

*$Rank(B) \leq min\{2,1\}=1$


So we can conclude that 


*

*$Rank(AB) \leq min\{Rank(A),Rank(B)\}=min\{1,1\}=1$


We know that : $A_{2x1}*B_{1x2}=AB_{2x2}$.
Therefore $Rank(AB) \neq 2 \implies$ singluar.
A: A is a 2×1 matrix and B is a 1×2 matrix,
$Rank(A)≤min\{$$2,1$}$=1$ 
$Rank(B)≤min\{$$2,1$}$  =1$
$Rank(AB)≤min\{$$Rank(A)$,$Rank(B)$}$=1$
$Rank(AB)<2$,not full rank,so not invertible.
Hence $A$ is singular.
A: In the case of rank one or zero matrices, it is easy to compute the eigenvalues and the corresponding eigenvectors directly. 
If $A = 0$, then $AB = 0$ which is clearly a singular matrix. Therefore suppose $B \neq 0$.
The eigenvalues of $AB$ are $BA$, $0$ and the corresponding eigenvectors are $A$, $B^\perp$. The zero eigenvalue implies the matrix $AB$ is singular.
A: Any vector in the range of $AB$ is in the range of $A$. But since $A$ is $2\times1$, the range of $A$ is just the span of its one column. So the range of $A$ is at most one-dimensional. So the range of $AB$ is at most one-dimensional.
(Note OP has changed the original problem. It now states $A$ is $2\times1$ but earlier stated $A$ is $1\times2$.)
A: If $A$ or $B$ is zero, then $AB$ is zero and therefore singular. 
Now suppose $A$ and $B$ are both non-zero. If $B = [b_1\ b_2]$, then the columns of $AB$ are $b_1A$ and $b_2A$. Therefore 
$$\operatorname{rank}(AB) = \dim\operatorname{Col}(AB) = \dim\operatorname{span}\{b_1A, b_2A\} = \dim\operatorname{span}\{A\} = 1 < 2,$$
so $AB$ is singular.
More generally, given $u \in \mathbb{R}^m$ and $v \in \mathbb{R}^n$, the $m\times n$ matrix $uv^T$ is called the outer product of $u$ and $v$. If $u$ and $v$ are non-zero, then $uv^T$ has rank one and every rank one matrix arises this way. Furthermore, the rank of a matrix $A$ is the smallest integer $k$ such that $A$ can be expressed as a sum of $k$ outer products.
