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?
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, $Ab_i = 0$.