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Let $A$ be a matrix, and $L$ be a linear mapping from $\mathrm{Row}(A)$ to $\mathrm{Col}(A)$ defined by $L(x) = Ax$. Prove $L$ is isomorphism.

I am having trouble to start this question. I know that the $\mathrm{Row}(A) = {}$orthogonal complement of $\mathrm{Null}(A)$, which I assume that this would be the starting point?

So if $x$ was an element of $\mathrm{Row}(A)$, then $x$ would also be an element of the orthogonal complement of $\mathrm{Null}(A)$, so somehow this would imply that $Ax = 0$ iff $x = 0$, which means the $\mathrm{Ker}(L) = \{0\}$ and thus be one-to-one? Does this sound like the right path? I'm not exactly sure how to prove it though

And then I am not sure how to prove onto for this case

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    I'm voting to close this question as off-topic because OP replaced the question with a random comment. – 2017-02-13
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    I rolled it back, since it has a nice answer and all. Just waiting for approval. – 2017-02-14

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I guess you define $\operatorname{Row}(A)=\operatorname{Col}(A^T)$, otherwise $Ax$ wouldn't make sense.

Let $x\in\operatorname{Row}(A)$ and suppose $L(x)=0$. Then $x\in\operatorname{Null}(A)\cap\operatorname{Row}(A)=\{0\}$, because $\operatorname{Row}(A)$ is the orthogonal complement of $\operatorname{Null}(A)$. Thus $x=0$ and the map is injective.

Since $\dim\operatorname{Row}(A)=\dim\operatorname{Col}(A)$ (the rank of $A$), the map is an isomorphism.

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    @SomeGuyGus Yes, you're right: it doesn't matter. Fixed. – 2017-02-13