If $\dim({\rm Im}(T))$ and $\dim({\rm Ker}(T))$ are finite $\Longrightarrow$ $\dim(V)$ is finite

linear algebralinear-transformations

Let be $F$ a field, $V$ and $W$ are vector spaces over $F$. Prove that
if it exist a linear tranformation $T:V\rightarrow W$ with
$\dim({\rm Im}(T))$ and $\dim({\rm Ker}(T))$ both finite, then $\dim(V)$ is finite.

What we know is that:

  • There exists a finite set of vectors $\left \{ v_1,…,v_n \right \}$ such that generates ${\rm Ker}(T)$
  • There exists a finite set of vectors $\left \{ w_1,…,w_m \right \}$ such that for each one exists $u_i \in V$, such that $T(w_i)=u_i$

Since there exists a finite quantity of $u_i \in V $ such that $T(w_i)=u_i$, and $\dim({\rm Ker}(T))<\infty$, then $\dim(V)$ must be finite. But I'm not sure about how can I conclude, can you help me please?

Best Answer

Hint: proceed as in the usual proof that $\dim V = \dim \ker T + \dim \text{im} \ T$, that is, consider a basis $B_0$ of $\ker T$ and extend it to a basis $B$ of $V$. Now, show that the image of $B \setminus B_0$ is a basis for $\text{im} \ T$.

These remarks and the hypothesis tells us that both $B_0$ and $B\setminus B_0$ must be finite, hence the basis $B$ of $V$ is finite. This completes the proof.

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