1
$\begingroup$

This is the problem 12 in Chapter 3.A from Axler's book "Linear Algebra done right". I need someone to check if my proof is good. I also written my thought process alongside what i've written on a paper.

This is the solution:

Let $L_1, L_2, ..., L_m \in L(V, W)$, where $dimV = n > 0$, and $dimW = \infty$.

Linear combination of linear maps is defined as $ \left(\sum_{i=1}^{m}a_iL_i \right)(v) = \sum_{i=1}^{m}(a_iL_i(v))$, and $v\in V$.

We need to show that there exist a sequence of linear maps $L_1, ..., L_m$ which is independent for every finite $m$.
Since $W$ is infinite dimensional, there exist a sequence $w_1, ..., w_m$ which is linearly independent for every finite $m$.

If we look at the right side of the definition for linear combination, we see that those are vectors in $W$. If we cut put there $w_i$ instead of $L_i(v)$, we could solve the problem. We can express $v$ as $v = \sum_{j=1}^{n}(c_jv_j)$, where $v_1, v_2, ..., v_n$ is a basis for V. Now, $$\sum_{i=1}^{m}(a_iL_i(v)) = \sum_{i=1}^{m}(a_iL_i(\sum_{j=1}^{n}(c_jv_j))) = \sum_{i=1}^{m}(a_i(\sum_{j=1}^{n}c_jL_i(v_j))).$$ If we could convert that inner sum into $w_i$, problem is solved. Good way to remove such sum is to define $L_i(v_j) = 0$, when $j \neq 1$, and $L_i(v_1) = w_i$.
Now our sum above becomes $$\sum_{i=1}^{m}(a_i(\sum_{j=1}^{n}c_jL_i(v_j))) = \sum_{i=1}^{m}(a_ic_1w_i).$$

Since $w_1, ..., w_m$ is independent list, $\sum_{i=1}^{m}(a_ic_1w_i) = 0 \rightarrow (a_1c_1 = ... = a_mc_1 = 0)$. This means that either $c_1 = 0$, or $a_1 = a_2 = ... = a_m = 0$, but since this must hold for every vector in $V$ (even those vectors where $c_1 \neq 0$), we conclude that $a_1 = a_2 = ... = a_m = 0$. This completes the proof.

  • 0
    The very claim is false: Check what happens if $\dim V=0$. – 2017-02-05
  • 0
    I forgot to mention that $dimV > 0$, because otherwise, $V = \{0\}$, and every linear map on $V$ must map only to $0$ :D – 2017-02-05

1 Answers 1

1

Yes, it's basically fine. However, I'd make it straighter.

Let $(w_k)$ be a sequence of vectors in $W$ such that $\{w_1,\dots,w_m\}$ is linearly independent, for each $m\ge1$.

Let $\{v_1,\dots,v_n\}$ be a basis for $V$. Define $L_k\colon V\to W$ by $$ L_k(v_1)=w_k, \quad L_k(v_i)=0\ (i=2,\dots,n) $$ Now we can easily prove that, for each $m$, $\{L_1,\dots,L_m\}$ is linearly independent. Indeed, if $a_1L_1+\dots+a_mL_m=0$, then $$ a_1L_1(v_1)+\dots+a_mL_m(v_1)=0 $$ that is, $$ a_1w_1+\dots+a_mw_m=0 $$ and therefore $a_1=a_2=\dots=a_m=0$.