My thought is that since a matrix can only be surjective if null(T)= n-m, no matrix can have a nullity of -1 therefore a 5x4 matrix cannot be surjective. Is this a valid proof?
$\endgroup$ 12 Answers
$\begingroup$A $5×4$ matrix represents a linear transformation $T$ from a $4$ dimensional vector space to a $5$ dimensional vector space. If $T$ is surjective, then the rank of $T=5$. But rank nullity theorem implies that $\operatorname{rank}(T)+\operatorname{nullity}(T)=4$, i.e. $\operatorname{nullity}(T)=-1,$a contradiction.
$\endgroup$ $\begingroup$HINT
Recall that a basis for $\mathbb{R^5}$ requires exactly $5$ linearly independent vectors.
What is the maximum rank for $A\in M_{5\times 4}$?
$\endgroup$ 7
