I am trying to solve this exercise:
Let A be a matrix of size $n \times n$ and X a non-zero column matrix of size $n.$ Show that if there exists a matrix $X$ such that $AX = O_n,$ then $A$ is not invertible.
If I understood correctly, I have to show that $AX = O_n \implies$ that $A$ is not invertible, i.e. that there is no matrix $B$ such that $AB = I_n.$
I don't have much idea how to solve this problem, which way to go, so I tried to develop this product a little bit and I found : $$AX =x_1 \pmatrix{a_{11} \\ \vdots\\ a_{n1}} + x_2 \pmatrix{a_{12} \\ \vdots\\ a_{n2}} + \cdot\cdot\cdot + x_i \pmatrix{a_{1i} \\ \vdots\\ a_{ni}} + \cdot\cdot\cdot + x_n \pmatrix{a_{1n} \\ \vdots\\ a_{nn}}$$
From here (I know it's not much ^^), I haven't found any ideas to show what I want. I can write a sum like $x_1 \cdot a_{11} + ... + x_n \cdot a_{1i} = 0$, but I don't see how this can be useful.
Thanks in advance for your help / advice!
We can assume, by contradiction, that $A$ is invertible and $B$ is its inverse. Then $$X=BAX=BO_n=O_n,$$ which contradicts the assumption $X\neq O_n.$