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

Question

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 05 '15 at 01:36
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. — Saúl Pilatowsky-Cameo, Apr 05 '15 at 01:48
wikipedia might help you already http://en.wikipedia.org/wiki/Fundamental_theorem_on_homomorphisms — RealAnalysis, Apr 05 '15 at 01:52
Thanks! But this seems too complicated, there must be a simpler way... — Saúl Pilatowsky-Cameo, Apr 05 '15 at 01:56
@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 05 '15 at 01:58
V/Ker(T) is a quotient gruop right? I really haven't got many experience working with those, this is my first linear algebra course, and I haven't taken any abstract algebra. — Saúl Pilatowsky-Cameo, Apr 05 '15 at 02:08

score 2 · Accepted Answer · edited Nov 29 '18 at 15:49

2

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.

edited Nov 29 '18 at 15:49

amWhy

209,954

answered Apr 05 '15 at 02:18

zhw.

105,693

Sounds right! But how do you conclude that the image of the span of ${v_1, \dots ,v_n}$ has dimension $n$? – Saúl Pilatowsky-Cameo Apr 05 '15 at 02:47
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 05 '15 at 03:12
Of course, thank you! – Saúl Pilatowsky-Cameo Apr 05 '15 at 04:40

score 1 · Answer 2 · answered Nov 29 '18 at 16:15

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 $v-v'\in\ker(T)$.

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

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