If $T:V \to W$ is such that both $\ker(T)$ and $\operatorname{Im}(T)$ are finite-dimensional, then $V$ is finite-dimensional

Let $V,W$ be vector spaces. Prove that if there exists a linear transformation $T:V \to W$ such that both $\ker(T)$ and $\operatorname{Im}(T)$ are finite-dimensional then $V$ is finite-dimensional as well.

I'm not sure how to prove this. My first intuition was to use the dimension theorem, but I can't because it requires that the domain is finite dimensional, and that's what I want to prove.

• use that for any linear transformation we have that $V/Ker(T)$ is isomorphic to $Im(T).$ – RealAnalysis Apr 5 '15 at 1:36
• Well, that would solve the problem, but I would need to prove it... How can I prove this? I sense it requires more work than just prooving it more directly somehow. – Saulpila Apr 5 '15 at 1:48
• Thanks! But this seems too complicated, there must be a simpler way... – Saulpila Apr 5 '15 at 1:56
• @Saulpila: Prove it directly. There's a reasonably obvious map $V / \mathrm{Ker}(T) \to \mathrm{Im}(T)$, so prove it's an isomorphism. – Clive Newstead Apr 5 '15 at 1:58

Suppose $$V$$ is infinite dimensional. Let $$\{u_1, \dots, u_m\}$$ be a basis for $$\ker{(T)}$$. Then we can extend $$\{u_1,\dots ,u_m\}$$ to a linearly independent set $$\{u_1,\dots ,u_m,v_1,\dots , v_n\}$$, with $$n> \dim \text {Im}(T)$$. The image of the span of $$\{v_1, \dots ,v_n\}$$ has dimension $$n,$$ contradiction.
• Sounds right! But how do you conclude that the image of the span of $\{v_1, \dots ,v_n\}$ has dimension $n$? – Saulpila Apr 5 '15 at 2:47
• Suppose $\sum a_kT(v_k) = 0.$ Then $T(\sum a_kv_k)= 0.\implies \sum a_kv_k \in \text {Ker}(T).$ That happens iff all $a_k=0.$ – zhw. Apr 5 '15 at 3:12
Pick $$\{v_1,\dots,v_n\}$$ such that $$\{T(v_1),\dots,T(v_n)\}$$ is a spanning set for $$\operatorname{Im}(T)$$ and prove that, if $$\{u_1,\dots,u_m\}$$ is a spanning set for $$\ker(T)$$, then $$\{u_1,\dots,u_m,v_1,\dots,v_n\}$$ is a spanning set for $$V$$.
Hint: take $$v\in V$$; then $$T(v)=\sum_{i=1}^n\beta_iT(v_i)$$ and, if we set $$v'=\sum_{i=1}^n\beta_iv_i$$, we have $$T(v)=T(v')$$, so $$v-v'\in\ker(T)$$.