Why if $B$ isn' t invertible then there exists $x\neq 0$ such that $Bx=0$?

Question

If $B$ isn' t invertible then there exists $x\neq 0$ such that $Bx=0$.

So $B$ would be $0$? Why?

Answer 1

Many ways to do this. For instance, $B$ is non-invertible $\implies$ $b_1,b_2\dots b_n,$the columns of $B,$ are linearly dependent. This means there exists $c_i$ scalars such that $c_1b_1+c_2b_2+\dots c_nb_n = 0.$ Now take $x = (c_1,c_2,...c_n)^T\neq 0.$

Answer 2

Example: $B=\begin{bmatrix} 1 & 0 \\ 0 & 0 \\ \end{bmatrix}$. With $x=(0,1)^T$ we have $Bx=0$, but $B \ne 0.$