What is the intuition behind fundamental theorem of linear maps

Question

I am self-studying Linear Algebra Done Right. I follow the proof in the book on Fundamental Theorem of Linear Maps.

$$\text{dim }V = \text{dim null }T + \text{dim range } T$$

My understanding is that, the basis of null $T$ with length $m $ can be extended to a basis of $V$ with length $m+v$. And then we can show that dim range $T = n$.

But I don't see the intuition/ picture of this. Why the dimension of $V$ is the sum of dimension of its null space and the range of $T$?

Answer 1

Take a basis $\{n_1,\ldots,n_k\}$ of $\operatorname{null}T$. Let $\{v_1,\ldots,v_l\}\subset V$ be such that $\{n_1,\ldots,n_k\}\cup\{v_1,\ldots,v_l\}$ is a basis of $V$. Then $\bigl\{T(v_1),\ldots,T(v_l)\bigr\}$ is a basis of $\operatorname{range}T$ and therefore$$\dim\operatorname{range}T=l=k+l-k=\dim V-\dim\operatorname{null}T.$$

Answer 2

What seems like the "intuition" to me involves concepts you may not have studied yet.

Say $T:V\to W$ and $Z$ is the nullspace of $T$. Now it's clear that $T$ "induces" a map $$\tilde T:V/Z\to W$$ in a "natural" way. Since we factored out the nullspace of $T$ it's clear that $\tilde T$ is injective, so it's clear that the dimension of the range of $\tilde T$ is equal to $\dim(V/Z)=\dim(V)-\dim(Z)$. And it's clear that the range of $\tilde T$ is the same as the range of $T$.

Answer 3

If anyone is still looking for it, the way I like to think about the intuition is that a linear map sends vectors in its domain to a certain space in its codomain. In the process, some vectors may be "lost", so that the dimension of the range is smaller than the dimension of the domain. How much smaller? That's measured by how many vectors are sent to zero.

That is, the dimension of the null space of T measures how many dimensions are lost when performing the transformation on V, and the dimension of the range of T is how many dimensions stay around. Together they add up to the dimension you started with.