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

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

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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.

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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.

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