So, here's a corollary that I'm trying to prove using the theorem. I've already proved it and my proof has been verified in a previous post that I made.

Let $f: V \to W$ be a linear map such that $\dim(V) = \dim(W)$. Then, $f$ is surjective if and only if it is injective.

Proof Attempt:

Let $f$ be surjective. Then, $f(V) = Im(f) = W$. Then:

$\dim(Ker(f)) + rank(f) = \dim(V)$

$\implies \dim(Ker(f)) + \dim(W) = \dim(V)$

$\implies \dim(Ker(f)) = 0$

Now, let $v_1, v_2 \in V$ such that $f(v_1)=f(v_2)$. Then:

$f(v_1)-f(v_2) = 0$

$\implies f(v_1-v_2) = 0$

$\implies v_1-v_2 \in Ker(f)$

$\implies v_1 = v_2$

That proves injectivity.

Now, suppose that $f$ is injective. Then, we need to show that $f(V) = W$. Since $f$ is injective, it follows that $Ker(f) = \{0\}$ and $\dim(Ker(f)) = 0$. So, we have:

$rank(f) = \dim(Im(f)) = \dim(V) = \dim(W)$.

Let $\alpha$ be the basis of $W$ and $\beta$ be the basis of $Im(f)$. Clearly, the basis lengths are the same based on the analysis above. Now, $\beta$ is a basis for $W$ by a trivial application of the Basis Extension Theorem. So:

$L(\beta) = Im(f)$

$L(\beta) = W$

$\implies W = Im(f)$

And that proves surjectivity. This proves the desired result.

The only part where I'm a bit hesitant about the quality of my argument is the last portion. I'm not quite sure how I can phrase it in a better way. Does my proof above work or no? If so, how can I fix it?


Yes, it's correct, provided the spaces are finite dimensional.

In the last part, you are proving ${\rm rank}(f) =\dim W$ implies surjectivity of $f$.
Well, ${\rm im}(f)\subseteq W$ is a subspace, but $W$ has only one $\dim W$ dimensional subspace, itself (e.g. because any linearly independent set of $\dim W$ number of vectors is a basis of $W$, ultimately yes, using a basis exchange argument).

  • $\begingroup$ Thank you for the response $\endgroup$ – Abhi Feb 26 '20 at 19:05

The result that I believe you're showing in the last part is that if we have a vector space $A$, a space $B \subset A$ and if $dim(B) = dim(A)$, then $B=A$

This is true because if $dim(B) = dim(A)= n$ and we let $\beta:a_1,...,a_n$ is a basis for $B$, we then take $a \in A$ then either it is in the span or it isn't. If it's not then $v$ is linearly independent of $\beta$ which can't be true because $dim(A) = n$.

So if $a \in A$, then $a \in A$. Hence, $A=B$.

In this case, $Im(f) \subset W$ and $dim(Im(f)) = dim(W)$.

So then $Im(f) = W$ and so we can conclude that $f$ is surjective.

(Edited names of subspaces in example to avoid confusion)

  • $\begingroup$ Not at all. Let $f:V \to W$ be a linear map such that $dim(V) = dim(W)$. If $f$ is, then we need to show that $f(V) = W$ $\endgroup$ – Abhi Feb 26 '20 at 19:10
  • $\begingroup$ Excuse me if I'm misunderstanding but is showing that $f(V) = W$ not the same as showing that $Im(f) = W$? $\endgroup$ – iCaird Feb 26 '20 at 19:13
  • $\begingroup$ Oh no, you initially wrote that W = V (3rd line). But yes, that’s the same. $\endgroup$ – Abhi Feb 26 '20 at 19:14
  • 1
    $\begingroup$ Yeah sorry, using the $V$ and $W$ for my example is a bit confusing. $\endgroup$ – iCaird Feb 26 '20 at 19:16

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