<s>[INST] Given the matrices $A\in M_{m\times n}, B\in M_{n\times p}: \ B\neq 0$, suppose $AB=0$. Prove: there exists a non-trivial solution for $Ax=0$ Given two matrices: $A_{m\times n}$ and  $B_{n\times p}$, such that $B\neq 0$,
suppose $AB=0$.
Prove: there exists a non-trivial solution for $Ax=0$. 
My Attempt:
suppose by contradiction that the only solution for the linear equation system $Ax=0$ is the trivial solution, meaning $Ax=0 \iff x=0$. 
$$Ax=0\Longrightarrow\left[\begin{array}{cccc}
a_{11} & \cdots & \cdots & a_{1n}\\
\cdots & \cdots & \cdots & \cdots\\
\cdots & \cdots & \cdots & \cdots\\
a_{m1} & \cdots & \cdots & a_{mn}
\end{array}\right]\left[\begin{array}{c}
x_{1}\\
x_{2}\\
\vdots\\
x_{n}
\end{array}\right]=\left[\begin{array}{c}
0\\
\vdots\\
\vdots\\
0
\end{array}\right]$$
Hence, $$\left[\begin{array}{c}
a_{11}x_{1}+a_{12}x_{2}+\cdots+a_{1n}x_{n}=0\\
a_{21}x_{1}+a_{22}x_{2}+\cdots+a_{2n}x_{n}=0\\
\cdots\\
a_{m1}x_{1}+a_{m2}x_{2}+\cdots+a_{mn}x_{n}=0
\end{array}\right]$$
According to the assumption: $x_{1}=x_{2}=\cdots=x_{n}=0$
Now, Let $B=\left[\begin{array}{cccc}
b_{11} & \cdots & \cdots & b_{1p}\\
\cdots & \cdots & \cdots & \cdots\\
\cdots & \cdots & \cdots & \cdots\\
b_{n1} & \cdots & \cdots & b_{np}
\end{array}\right]$
Let $AB=(t)_{ij}$, as $\begin{array}{c}
1\le i\le m\\
1\le j\le p
\end{array}$
$AB=0\Longrightarrow\left[\begin{array}{cccc}
a_{11} & \cdots & \cdots & a_{1n}\\
\cdots & \cdots & \cdots & \cdots\\
\cdots & \cdots & \cdots & \cdots\\
a_{m1} & \cdots & \cdots & a_{mn}
\end{array}\right]\left[\begin{array}{cccc}
b_{11} & \cdots & \cdots & b_{1p}\\
\cdots & \cdots & \cdots & \cdots\\
\cdots & \cdots & \cdots & \cdots\\
b_{n1} & \cdots & \cdots & b_{np}
\end{array}\right]=\left[\begin{array}{c}
0\\
\cdots\\
\cdots\\
0
\end{array}\right]$
$\Longrightarrow\left[\begin{array}{cccc}
a_{11}b_{11}+\cdots+a_{1n}b_{n1}\  & \cdots\cdots & \cdots\cdots & a_{11}b_{1p}+\cdots+a_{1n}b_{np}\\
\cdots\cdots & \cdots\cdots & \cdots\cdots & \cdots\cdots\\
\cdots\cdots & \cdots\cdots & \cdots\cdots & \cdots\cdots\\
a_{m1}b_{11}+\cdots+a_{mn}b_{n1}\  & \cdots\cdots\  & \cdots\cdots\  & a_{m1}b_{1p}+\cdots+a_{mn}b_{np}
\end{array}\right]$
This means that for every $i,j:\ t{}_{ij}=0$, therefore $$\begin{array}{c}
a_{11}b_{11}+\cdots+a_{1n}b_{n1}=a_{11}x_{1}+a_{12}x_{2}+\cdots+a_{1n}x_{n}=0\\
\cdots\cdots\\
a_{m1}b_{11}+\cdots+a_{mn}b_{n1}=a_{m1}x_{1}+a_{m2}x_{2}+\cdots+a_{mn}x_{n}=0
\end{array}$$
but $x_{1}=\cdots=x_{n}=0$, then  $b_{11}=\cdots=b_{n1}=0$. 
this equity applies for every column of $B$, consequently, $B=0$.
But that's a contradiction to the fact that 
$B\neq 0$, then we'll get a contradiction to the assumption, therefore there exists a non-trivial solution for 
$Ax=0$, as wished.
It this correct?
 [/INST] If $B = [b_1,...,b_p] \neq 0$, there is some $i \in \{1,...,p\}$ such that $b_i \neq 0$. Then, by definition of the matrix product, $Ab_i = 0$.
</s>