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There are a couple other answers for this question here, but I'm actually looking for a proof verification (or really a foot foward for my own take in a different direction as the others).

My proof (so far):

Choose a basis $u_1,\dots,u_n$ of range $T$. Then for $v\in V$, we have $Tv=\lambda_1u_1+\dots+\lambda_n u_n$. Since $Tv\in$ range $T\subseteq V$, we have that $Tv$, an arbitrary element in $V$, is written in terms of a basis of range $T$, which is finite.

Assuming this is correct, what does it tell us about $\dim V$? A better question might be: is it possible to move on in the right direction from here?

My tired self wants to say that we would have $V=$ range $T$ (implying $\dim \ker T=0$ i.e. $T$ injective?), but I also feel like that's... wrong, to say the least.

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galois
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  • Do you think $V / \ker T$ is bijective with the image of $T$? You seem to be moving in that direction. – Eric Towers Apr 18 '16 at 02:53
  • Actually, while we learned that earlier this semester, I'm going back through homework and at this point in the book, V/ker T (assuming you mean a quotient space) hasn't been introduced yet. – galois Apr 18 '16 at 02:55

2 Answers2

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It is possible to move on in the correct direction, if you adjust things slightly.

Choose a basis $Tu_1,\ldots, Tu_n$ of the range, $T$. Then for $v\in V$ we have $Tv=\lambda_1Tu_1+\cdots + \lambda_nTu_n=T(\lambda_1u_1+\cdots+\lambda_nu_n)$.

Therefore $Tv-T(\lambda_1u_1+\cdots+\lambda_nu_n) = T(v-\sum_i \lambda_iu_i)= 0$. And so $(v-\sum_i\lambda_iu_i)\in\ker T$. Can you finish it from here?

jgon
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  • The last line implies that $v=\sum\lambda_i u_i$. We know that $\lambda_1u_1,\dots,\lambda_n u_n$ is linearly independent. Does it follow that it spans $V$, and is thus a basis of $V$ itself? – galois Apr 18 '16 at 03:00
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    @jaska, that's not quite right. The last line is that $v-\sum_i\lambda_iu_i \in\ker T$, it only implies $v=\sum_i\lambda_iu_i$ if $\ker T ={0}$. All we know is that $\ker T$ is finite dimensional. Say it has basis $w_j$. Then we know that $v-\sum_i\lambda_iu_i =\sum_j\mu_jw_j$. Then $v=\sum_i\lambda_iu_i+\sum_j\mu_jw_j$. – jgon Apr 18 '16 at 03:35
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    Then $v$ is in the span of the set ${u_i}\cup{w_j}$. Hence $V$ is spanned by a finite set, and is finite dimensional. – jgon Apr 18 '16 at 03:35
  • Couldn't we use the fact from my post that since $Tv$ is written in terms of a basis of range $T$, then $Tv\in span(u_1,\dots,u_n)$ means that an arbitrary value in $V$ is in the spanning set of a finite dimensional subspace? – galois Apr 18 '16 at 03:57
  • @jaska Honestly that's kind of what's going on in my post. You were missing some steps before because you were leaving out the degrees of freedom in $V$ that come from being able to move around in the kernel before taking the linear transformation. – jgon Apr 18 '16 at 03:59
  • Yeah, that's what I saw, but what I mean is: do you have to even touch the kernel? By showing that an arbitrary member of $V$ is represented as a basis of a finite dimensional subspace, isn't that enough? – galois Apr 18 '16 at 04:21
  • @jaska, but you haven't done that, you showed $Tv$ could be written in that form, not that $v$ could. – jgon Apr 18 '16 at 05:21
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Let $\beta=\{u_1,u_2,\ldots,u_m\}$ be a basis for ${\rm range}(T)$ and $\gamma=\{v_1,v_2,\ldots,v_n\}$ be a basis for ${\rm ker}(T)$. Then we may write $u_i=Tw_i$ for some $w_i\in V$, where $1\le i\le m$. Now, it suffices to claim that \begin{align} \alpha=\{v_1,v_2,\ldots,v_n,w_1,w_2,\ldots,w_m\} \end{align} is a basis for $V$.

  1. Given $a_1,a_2,\ldots,a_n$ and $b_1,b_2,\ldots,b_m$ be scalars such that \begin{align} \sum_{i=1}^na_iv_i+\sum_{j=1}^mb_jw_j={\it 0}.\tag{1} \end{align} Then \begin{align} {\it 0} &=T\left(\sum_{i=1}^na_iv_i+\sum_{j=1}^mb_jw_j\right)\\ &=\sum_{i=1}^na_iTv_i+\sum_{j=1}^mb_jTw_j\\ &=\sum_{j=1}^mb_ju_j, \end{align} which follows that $b_j=0$ for each $j$ because $\beta$ is a basis for ${\rm range}(T)$. Thus equation $(1)$ becomes $\sum_{i=1}^na_iv_i={\it 0}$, but since $\gamma$ is a basis for ${\rm ker}(T)$, we must have $a_i=0$ for each $i$. Hence $\alpha$ is linearly independent.
  2. Given $v\in V$, then $Tv\in{\rm range}(T)$ and we can write $Tv=\sum_{i=1}^m\lambda_iTw_i=T\left(\sum_{i=1}^m\lambda_iw_i\right)$ for some scalars $\lambda_1,\lambda_2,\ldots,\lambda_m$. It follows that $$T\left(v-\sum_{i=1}^m\lambda_iw_i\right)=Tv-T\left(\sum_{i=1}^m\lambda_iw_i\right)={\it 0},$$ that is, $v-\sum_{i=1}^m\lambda_iw_i\in{\rm ker}(T)$. Thus we can write $v-\sum_{i=1}^m\lambda_iw_i=\sum_{j=1}^n\mu_jv_j$ for some scalars $\mu_1,\mu_2,\ldots,\mu_n$. That is, $$v=\sum_{i=1}^m\lambda_iw_i+\sum_{j=1}^n\mu_jv_j.$$ Hence $\alpha$ generates $V$.

By checking the above two steps, we conclude that $\alpha$ is a basis for $V$, and therefore $V$ is finite-dimensional.

Solumilkyu
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