The sum of the dimensions of the kernel and image set equals the dimension of the vector space

Question

So, I have to prove the following result that I found in my book:

Let $f:V \to W$ be a linear map and let $V$ be a finite dimensional vector space. Then:

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

---

Proof Attempt:

Let $\dim(V) = n$. Since $Ker(f)$ is a subspace of $V$, it is finite-dimensional so let us have $\dim(Ker(f)) = m$.

Let $A = (v_1,v_2,...,v_n)$ be a basis of $V$ and let $B = (u_1,u_2,....,u_m)$ be a basis of $Ker(f)$. Let $w \in Im(f)$. Then, there exists a vector $v \in V$ such that $f(v) = w$. Since any vector $v$ can be written as a linear combination of the vectors in $A$, it is the case that every vector in $Im(f)$ can be written as a linear combination of the vectors $(f(v_1),....,f(v_n))$. This shows that Im(f) must be finite-dimensional since there are a list of vectors, linearly independent or not, that do span it.

Now, $B$ forms a linearly independent list of vectors in $V$ and $L(A+B) = V$, where $A+B$ simply means that we are combining both lists of vectors together.

By the Basis Extension Theorem, we can take vectors from $A$ and adjoin them to $B$ in order to form a list, called B', that is a basis for $V$. In particular, we need $n-m$ vectors from $A$ in order to do this. So, we have our new list written as follows:

$B' = (u_1,....,u_m,v_{j_1},....v_{j_{n-m}})$

Since any vector in $V$ can be written as a linear combination of the vectors in $B'$, it follows that any vector in $Im(f)$ can be written as a linear combination of the vectors $(f(u_1),....,f(u_m),f(v_{j_1}),.....,f(v_{j_{n-m}}))$. In other words, the linear hull of these vectors is equal to $Im(f)$.

However, if we consider any vector $v$ that is formed by the linear combination of the vectors in $B'$ and act on it with $f$, then we are left with:

$f(v) = \sum_{k=1}^{n-m} \beta_k v_{j_k}$

The reason for that is because $\forall i \in \{1,...,m\}: f(u_i) = 0$. So, we have a list of length $n-m$ that spans $Im(f)$ and is also a linearly independent list. Hence, it forms a basis for $Im(f)$. So, we have:

$rank(f) = n - m = \dim(V) - \dim(Ker(f))$

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

That proves the desired result.

I need feedback on the proof I have above. Is it correct? If it is, is there a shorter way to approach the problem? If it isn't, what's the exact flaw and what can I do to fix it? Is the proof clear or have I been too ambiguous in certain instances?

Answer 1

By Isomorphim's Theorem we have that $V/Ker f = Im f$. So we have that $dim(V/Ker f)=dim(Im f)$. But we have that $dim(V/ker f)= dim(V)-dim(ker f)$. So we have that $dim(V)=dim(ker f)+dim(Im f)$.

Answer 2

long comment

Use the conventional notation:

Let $\mathbb F$ be a vector space over the field $\mathbb F$. You should write that (I tell you this because most of the users here use different notation): $$\mathbb F:=\{(u_1,\ldots,u_n):u_i\in\mathbb F,\forall i\in\{1,\ldots,n\}$$ Then: We can have a linearly (in)dependent set $S$ and a vector space $V$ s.t. $\operatorname{span}(S)=V$ $S$ can be an extended base and therefore, linearly dependent, but only the base vectors span the vector space. So,ofcourse, your argument holds, according to the base Extension theorem. I would just shorten your proof on terms of reducing the text and involving more symbols because it would be readable. This ist just my own view, not an answer. But, all that can be found on other posts. I can add some of my literature to compare it for our own good. I'll be glad to discuss more.