Solution 3b 16

Suppose there exists a linear map on $V$ whose null space and range are both finite-dimensional. Prove that $V$ is finite-dimensional.

(Note 3.22 doesn't apply here since it assumes $V$ is finite-dimensional. 2.34 Also can't be used till we know $V$ is finite.)

Let $T$ be our linear map. Let $v_1,\dots,v_m$ be independent vectors in $V$ such that $Tv_1,\dots,Tv_r$ is a basis for $\text{range }T$ and $v_{r+1},\dots,v_m$ is a basis for $\text{null }T$ .

Let $v \in V$ , since $Tv_1,\dots,Tv_r$ is a basis for the range we can write

Tv = a_1Tv_1 + \dots + a_rTv

Subtract the right hand side and use linearity to get

T(v - a_1v_1 - \dots - a_rv_r) = 0

Since $v_{r+1},\dots,v_{m}$ is a basis for the nullspace we can write

v - a_1v_1 - \dots - a_rv_r = a_{r+1}v_{r+1} + \dots + a_mv_m

Rearranging gives

v = a_1v_1 + \dots + a_mv_m

Implying $\text{span}(v_1,\dots,v_m) = V$ meaning $V$ is finite-dimensional.

It isn't needed for the question but we can also show $v_1,\dots,v_m$ is linearly independent with a similar trick of tactically applying $T$

\begin{aligned} &a_1v_1+\dots+a_mv_m = 0 \\ &\implies a_1Tv_1+\dots+a_rTv_r = 0 \\ &\implies a_1=\dots=a_r=0 \\ &\implies a_{r+1}v_{r+1} + \dots + a_mv_m = 0 \\ &\implies a_{r+1} = \dots + a_m = 0 \end{aligned}

Corralary: We can replace the " $V$ is finite" assumption with "null $T$ and range $T$ are finite" in 3.22.

The first time I did this exercise it was really hard (took more then an hour). When an exercise is hard I let myself relax my mental rigor requirement and work backwords from "why is this intuitively true" to a proof, filling in holes as I go.

I don't think I would have solved this without knowing every vector can be represented as a sum of range and null components (which I learned in 1806). I worked backwards from there

The second time (now) the proof was totally obvious and I was able to clean it up a good bit. I guess I've improved :)