Endomorphisms on finite dimensional vector spaces $f: V \to V$ are surjective $\iff $ injective 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.  
 A: Just to give an actual answer to this question (some of which was already provided in comments):
It is not true that two vector spaces of the same dimension need be the same.
E.g. the space of linear polynomials $\{ a + b x \, | \, a , b \in \mathbb R\}$
is not literally "the same" as the space $\mathbb R^2$.
They are isomorphic (any two vector spaces over a given field of the same dimesion are isomorphic), but not the same space.
However, if $V$ is a finite-dimensional and $W \subset V$ is a subspace
of the same dimension of $V$, then $W = V$.  

The above situation is illustrative of something that happens quite often when we try to study more structured objects (vector spaces, fields, groups) in terms of less strucured invariants (such as numbers).  Typically, two objects having the same invariant will not be the same.  (E.g. there are different, indeed even non-isomorphic, groups of order $4$, there are different, even non-isomorphic,
degree two extensions of the field $\mathbb Q$, and so on.)  However,
often if we apply the invariant in the context of an inclusion, we go from equality of the invariant to equality of the actual structures.
E.g. if $H \subset G$ is an inclusion of finite groups, and $H$ and $G$ have the
same order, then $H = G$.
A: 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.
