Similar questions: Linear map $f:V\rightarrow V$ injective $\Longleftrightarrow$ surjective
Proposition: Let $V$ be a finite dimensional vector space over an arbitrary field $\mathbb{K}$. If $f: V \to V$ is an endomorphism of $V$, then $f$ is injective if and only if $f$ is surjective.
I want to understand (note: understand, not rigorously proof) that proposition using the Rank Theorem:
Rank Theorem: Let $V,W$ be finite vector spaces over a field $\mathbb{K}$ and let $f: V \to W$ be a linear mapping, then: $$\dim\ker f + \dim\text{im}f=\dim V $$
I want to use this approach because it seems to be something I could easily remember and reason out for myself in times of doubt, but there is in particular one step in this demonstration that I don't manage to wrap my head around and bothers me. I will comment on it.
My Approach: Given the Rank Theorem as above I choose $f$ to be surjective. This means that $\text{im}f= V \implies \dim \text{im}f=\dim V (**)$. So it follows immediately that $ \dim \ker f=0$. I want to show that $\ker f= \lbrace 0 \rbrace$.
I reason this as follows for myself: $\dim \ker f=0$ means that the dimension of the $\ker f$ is zero, the dimension of a point is zero or more generalized, the dimension of any singleton is zero, but since $\ker f$ is a subspace of $V$ it can only include the zero vector, therefore $\ker f = \lbrace 0 \rbrace$. I hope this reasoning is correct. (If not please comment/correct)
This would show that if $f$ is surjective then $f$ is injective.
Now for the other side (where I am having troubles with) I set $f$ to be injective, this means that $ \ker f= \lbrace 0 \rbrace \implies \dim \ker f =0$. Again I make use of the rank theorem and I obtain that $$\dim \text{Im}f=\dim \tag{!}V$$ And this is where I am stuck. While I was looking for similar approaches on the internet I found that most people stop here and use the above identity to conclude the proof. However, to be sincere, the above equation bothers me since the way I interpret it doesn't finish the proof.
In my approach above I made use of the following: $$\text{im}f= V \implies \dim \text{im}f=\dim V \tag{**} $$ and I understand this side of the implication, if two vector spaces are equivalent, then so is their dimension. However to complete the proof I would need the other side: $$\text{im}f= V \Longleftarrow \dim \text{im}f=\dim V \tag{**''}$$ And I cannot wrap my head around this one. If the dimension of two vector spaces are the same, then the vector spaces are the same? I am only a freshman to the subject but I doubt that this can be true for all of Mathematics, and if it is then I clearly missed the theorem of that statement.
I am missing a good reasoning here why the above implication holds true and I am somehow sure it has something to do with $f$ being an endomorphism, but I cannot find it. I would appreciate help to formulate out that missing step.
Your argument about $\dim\ker f=0$ is a bit convoluted, but, well, correct. If you understand it best that way, I'm fine with it. Another way: If there wee any nonzero vector $v$ in $\ker f$, then the span of $v$ must have (at least) dimension $1$.
Your are sticking your finger right into the spot! Equality of dimension of a subspace with the whole space does not in general imply equality of the spaces. IN fact, this is the point where the argument breaks in case of infinite-dimensional spaces. However, in the finite case: Let $U\subsetneq V$ be spaces with $\dim U=\dim V=n<\infty$. Pick $u\in U\setminus V$. Then $u$ is linearly independent from $U$, i.e. together with a basis of $U$ we obtain $n+1$ linearly independant vectors in $V$. These can be completed to a basis of $V$ so that $\dim V\ge n+1$. In short: A proper subspace of a finite-dimensional vector space has strictly smaller dimension.