# Corollary of The Rank-Nullity Theorem

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).

• Thank you for the response – 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)

• 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 $f(V) = W$ – Abhi Feb 26 '20 at 19:10
• Excuse me if I'm misunderstanding but is showing that $f(V) = W$ not the same as showing that $Im(f) = W$? – iCaird Feb 26 '20 at 19:13
• Oh no, you initially wrote that W = V (3rd line). But yes, that’s the same. – Abhi Feb 26 '20 at 19:14
• Yeah sorry, using the $V$ and $W$ for my example is a bit confusing. – iCaird Feb 26 '20 at 19:16