Image of an injective linear transform is linearly independent

Question

Let $L:V\mapsto W$ be an injective linear transform and $U$ a linearly independent subset of the vector space $V$. Show that $L(U)$ is linearly independent.

Proof : Since $U$ is $LI$, then if $s \in U$ $$s_1x_1 + ...s_nx_n=0$$

has only the trivial solution. So every term $s_nx_n=0$, and since $L$ is linear we have that $L(s_nx_n)=L(0)=0$. From injectivity it follows that $$L(s_1x_1)=L(s_nx_n) \rightarrow s_1x_1=s_nx_n$$

Now each vector $s_nx_n$ is mapped into $0$, and so the image of the set $U$

$$L(U)=L(s_1x_1)+...L(s_1x_1)=L(s_1x_1+...s_nx_n)=0$$

when $x_i=0,(i=1,...,n)$. Therefore $L(U)$ is linearly independent. What is the significance of $U \subset V$? Other than the fact that since $V \mapsto W$ is injective, certainly this implies injectivity for elements of $U$ as well. Does this show that a $LI$ set is mapped onto another $LI$ set iff the linear transform is injective? (... if not injective then $L(s_1x_1)=L(s_nx_n) \rightarrow s_1x_1 \neq s_nx_n$ so we would have a non-trivial solution )

Answer 1

It's not enough to use that $L$ restricted to $U$ is injective, that is that $Ls_i$ are pairwise distinct for $s_i\in U$, since e.g. for $V=W=\Bbb R^2$ and $L(x,y) =(x+2y,x+2y)$, the standard basis $(1,0),\ (0,1)$ are mapped to different but linearly dependent (parallel) vectors.

You should start the proof by assuming $$Ls_1x_1+Ls_2x_2+\dots +Ls_nx_n=0$$ for some distinct elements $s_i\in U$ and $x_i\in\Bbb R$, and you should deduce all $x_i=0$, using that $s_i$ are linearly independent and that $\ker L=\{0\}$.

Answer 2

We need $U \subset V$ to say that $U$ is linear independent. Otherwise we wouldn't be able to form linear combinations on $U$. Also we need $U \subset V$ because our map $L$ maps from $V$ to $W$ and we want to prove something about $L(U)$, so we need to be able to apply $L$ to $U$.

Here is a proof of the claim:

Since $U$ is linear independent we have for vectors $u_i \in U$ for all $c_i \in \mathbb{F}$ with $$\sum_i c_i u_i = 0$$ that $c_i = 0 $ for every $i$.

Now we need to show that the set $L(U)$ is linear independent, so we look at elements $L(u_i)$. We need to show that for all $c_i \in \mathbb{F}$ with $$\sum_i c_i L(u_i) = 0$$ $c_i = 0$ for every $i$ holds. Since $L$ is linear we see that $$0 =\sum_i c_i L(u_i) = L \left(\sum_i c_i u_i \right) $$ holds. Now, since $L$ is injective we know that $L$ only sends $0$ to $0$, so $\sum_i c_i u_i = 0$ follows. And because of the linear independence of $U$ we conclude $c_i =0$ for every $i$ and thus $L(U)$ is linear independent.

Your iff-statement is wrong. Only one direction holds. There is an easy counterexample: Every subset $\{v\} \subset V$ with $v \neq 0$ is linear independent since it consists of only one element which is not zero. Now every linear map $L: V\to W$ with $L(v) \neq 0$ sends $\{v\}$ to a linear independent subset of $W$ but these maps don't have to be injective in general.

However if your subset $U \subset V$ is a basis of $V$ and gets send to a linear independent subset $L(U) \subset W$ of $W$, then $L$ is injective. We need to show $\ker(L) = \{0\}$. So suppose $L(v) = 0$. Since $U$ is a Basis we can write $ v = \sum_i c_i u_i$ for some $c_i \in \mathbb{F}$. This means $$ 0 = L(v) = L\left(\sum_i c_i u_i\right) = \sum_i c_i L(u_i) $$ and because $L(U)$ is linear independent it follows that $c_i = 0$ for every $i$ and as such $$v = \sum_i c_i u_i = \sum_i 0 u_i = 0$$ so $\ker(L) = \{0\}$ and we conclude that $L$ is injective.