Given rank and nullity, prove that $rank(B^2)=rank(B)$

Question

Let $B=\begin{bmatrix} 2 & 0 &4 \\ 1& -1&3 \\ 2& 1 & 3 \end{bmatrix}$

We have $rank(B)=2$, $nullity(B)=1$, $ker(B)\;\cap \; col(B)=\left \{ \mathbf{0} \right \}$

Prove, without computing $B^2$, that $rank(B^2)=2$

Answer 1

Hint: By the conditions, we have $\mathrm{col}(B)\oplus\ker(B) =\Bbb R^3$, and thus $B$ restricted to $\mathrm{col}(B)$ is injective.