# Showing a linear transformation is surjective given a minor of its coresponding matrix is invertible

I have been working practice problems that give conditions if a map is bijective, surjective, etc. and this could be considered a follow up to this post Condition on minors of a matrix to check when linear transformation is injective but I do not think they are related other than they are from the same family of similar facts I am trying to uncatalogued for a test. I

Let $V,W$ be two finite dimensional vectors spaces. Suppose that $f: V \rightarrow W$ is a linear mapping and let $A$ be the matrix corresponding to the linear map $f$.

How do you show if $\dim V \geq \dim W$ and there exists a minor of $A$ with the same of order as $\dim W$, then the map $f$ is surjective.

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Pick bases for $V$ and $W$ and let $A$ be the corresponding coordinate matrix of $f$.
If there is a minor of order $\mathrm{dim}(W)=m$, then the $m$ columns of $A$ for which this minor is non-zero, form an invertible $m \times m$ matrix (so in particular they are linearly independent). Let's say these columns correspond to ${\bf v}_1,\dots,{\bf v}_m$ in the basis for $V$. We have $\{ f({\bf v}_1),\dots, f({\bf v}_m) \}$ is a linearly independent set. Thus $m \leq \dim(f(V)) \leq \dim(W)=m$. Therefore, $\dim(f(V))=\dim(W)$.