Let $V$ and $W$ be vector spaces and let $T: V\to W$ be a linear transformation. Let $\{v_1,v_2,\ldots,v_p\}$ be a linearly dependent set of vectors in $V$. Show that $\{Tv_1,Tv_2,\ldots,Tv_p\}$ is also linearly dependent.
Here's what I have, I need someone to tell me if they think this works.
Since $\{v_1,\ldots,v_p\}$ is linearly dependent, we know that there are scalars that are not all zero (lets this scalar be an) that make $a_1v_1+\ldots+a_nv_n = 0$
Once we apply the transformation we get $T(a_1v_1+\ldots+a_nv_n)=T(0)$
Because are told that this thing is a linear transformation, we know its closed under addition and scalar multiplication. We can use this to change the set to.
$T(a_1v_1)+\ldots+T(a_nv_n) = T(0)$
and we can also pull those scalars out since its dependent.
$a_1T(v_1)+…+a_nT(v_n) = T(0)$
the scalars are still not all zero since we just factored them out of the transformation. Therefore, we have a set here that has scalars, not all zero that gives us $T(0)$ which is $0$. This set with the transformation applied is linearly dependent which is what we needed to show.
The only question I have here is about the end where I said that $T(0) = 0$. Can I make that assumption or am I missing something?
Best Answer
As noted in the comments, $T(0) = 0$ holds for any linear map:
$$T(0) = T(0 + 0) = T(0) + T(0) = 2T(0) \implies T(0) = 0$$
A bit easier is to show the contrapositive of your statement: if $\{Tv_1, \ldots, Tv_p\}$ is linearly independent, then $\{v_1, \ldots, v_p\}$ is also linearly independent.
Take scalars $\alpha_1, \ldots, \alpha_p$ such that $0 = \alpha_1 v_1 + \cdots + \alpha_p v_p$.
Apply $T$ on both sides:
$$0 = T(0) = T(\alpha_1 v_1 + \cdots + \alpha_p v_p) = \alpha_1 Tv_1 + \cdots + \alpha_p Tv_p$$
Therefore, $\alpha_1 = \ldots = \alpha_p = 0$, which shows that $\{v_1, \ldots, v_p\}$ is linearly independent.