Intuition behind the dimension theorem I am learning about the dimension theorem:
Let $L: V \to W$ be a linear map from the vector space $V$ to the vector space $W$. If $V$ is finite dimensional then $\dim \ker(L) + \dim \text{Range}(L) = \dim V$.
What is the intuition behind this result?
I find it counterintuitive that the dimension of a subspace from $V$ (null space of $V$) plus the dimension of a subspace in $W$ (range of $L$) add up to the dimension of $V$, the vector space where we are mapping from. Any insights are appreciated.
 A: You can think of the dimension of a vector space as the amount of information that it contains. In this way of thinking, if $L : V \to W$ then we can think of $\dim(\ker(L))$ as the amount of information destroyed when going from $V$ to $W$, and $\dim(\mathrm{range}(L))$ as the amount of information that is not destroyed. In this context, if $L$ doesn't destroy any information then we would get $\dim(\mathrm{range}(L)) = \dim(V)$. However, if some of the information is destroyed then we can think of this formula as stating something like: the amount of information destroyed ($\dim (\ker(L))$) plus the amount of information that is not destroyed ($\dim(\mathrm{range} (L))$) equals the original amount of information, which is $\dim(V)$.
A: Let $n=\dim\operatorname{Range}L$. Then there is a basis $\{w_1,\ldots,w_n\}$ of $\operatorname{Range}L$. Since each $w_k$ is in the range of $L$, there is some $v_k\in V$ such that $L(v_k)=w_k$ and, since $\{w_1,\ldots,w_n\}$ is linearly independent, so is $\{v_1,v_2,\ldots,v_n\}$. Now, if $m=\dim\ker L$, let $\{v_1',v_2',\ldots,v_m'\}$ be a basis of $\ker L$. It turns out that then $\{v_1,v_2,\ldots,v_n\}\cup\{v_1',v_2',\ldots,v_m'\}$ is a basis of $V$ (that's not hard to prove), and therefore$$\dim V=m+n=\dim\ker L+\dim\operatorname{Range}L.$$
