# Proving a matrix is surjective

Is a $$n \times n$$ matrix always surjective? If it is, how can I prove it using the rank-nullity theorem?

• Small language point for the future. Matrices are not surjective The linear functions they determine may be. – Ethan Bolker Nov 24 '18 at 23:15

No it's not. To see this consider the zero matrix as a map from $$\mathbb{R}^n\rightarrow \mathbb{R}^n$$ which is clearly not surjective.
Recall that surjective means that for any $$b\in \mathbb{R^n}$$ the system
$$Ax=b$$
has solution, that is true if and only if $$\dim(Im(A))=n$$ that if and only if $$A$$ is full rank.
Maybe your getting confused because there is a fact that says that a $$n\times n$$ matrix which has null kernel is always surjective. This is a classical applications of rank-nullity theorem because $$n=\dim Ker (A) + \dim Im (A)$$ and its clear that $$\dim Ker(A)=0 \implies \dim Im(A)=n$$ which in this case means that $$A$$ is surjective.