If $X$ is linearly independent, then $T(X)$ is also linearly independent? Here is my homework question,

Let $V$ and $W$ be vector spaces over a field $F$ and let $T$ from $V$ to $W$ be an isomorphism. Let $X$={$v_{1},v_{2},...,v_{n}$} be a subset of V, and recall that T(X)={$T(v_{1}),T(v_{2}),...,T(v_{n})$}.
a. Prove that if $X$ is linearly independent, then $T(X)$ is also linearly independent.
b. Prove that if $X$ spans $V$, then $T(X)$ spans $W$.

I am trying to solve a. but I don't know how to use isomorphism here...
My work is.. Suppose X is linearly independent. Then for all $a_{1}v_{1},a_{2}v_{2},...,a_{n}v_{n}$ = 0. So, $a_{1},a_{2},...,a_{n}$ = 0. but I can't go further...
 A: I understand that this is a homework question. However based on your previous question I am willing to give a little more help to help you get the ball rolling and basic ideas in these problems. I will help you out with $(a)$ because $(b)$ almost immediately follows from the definitions of $T$ being an isomorphism (in particular that it is surjective).
For $(a)$, you have an implication  $p \implies q$, where $p$ and $q$ are respectively the statements


$p$: $X$ is linearly independent
$q$: $T(X)$ is linearly independent.


If you don't know about the contrapositive yet, it is a powerful way of proving statements. In your context it says that proving $p \implies q$ is equivalent to proving $\neg q \implies \neg p$ where the "$\neg$" symbol is logical negation.
So suppose we know $\neg q$. That is, we assume that the vectors in the collection $T(X)$ are linearly dependent. Now you will out the following details:


1) What does it mean for the vectors in $T(X)$ to be linearly dependent? You just need to apply the definition of linear dependence.


After you have written that down, using linearity of $T$ this implies that (.....)? You can fill in $(\ldots \ldots )$ here by looking at question 2) below:


2) Is there a vector in the kernel of $T$ now? Proceed to 3) below.
3)  Using the fact that $T$ is an isomorphism can you conclude from here that $X$ is linearly dependent? Remember we started of with $\neg q$ and we want to prove $\neg p$, where $\neg p$ is the statement   "$X$ is linearly dependent."


$\textbf{Edit:}$ Here is my response to your solution below: You cannot start with saying " suppose that $a_1 = a_2 = \ldots a_n = 0$." This is non-sensical. In fact in your proof the only thing you've done is showing that $T$ maps zero to zero. This is because you started of with saying that all the $a_i's$ were zero, so that $T$ applied to any linear combination involving these $a_i's$ will be zero anyway. Do you see this does not lead to a proof of what you want to prove?
What you need to do is this: Suppose we have a linear combination of the $T(v_i)'s$ that gives us zero. Viz. there are scalars $a_1 , a_2, a_3, 
\ldots a_n$ such that
$$a_1T(v_1) + 
\ldots a_nT(v_n) = 0.$$
Remember: You want to conclude from here, using information about $X$, that this linear combination can only be the trivial linear combination.  Now by linearity of $T$ this means that the vector $a_1v_1 + 
\ldots a_nv_n$ is in the kernel of $T$. But then $T$ is an isomorphism so this means that $a_1v_1 + 
\ldots a_nv_n = 0$.
Do you see how to prove it correctly now?
Now because $X$ is linearly independent, this forces us to conclude that $a_1 = a_2 = \ldots = a_n = 0$.
A: To expand on Martin's Hint:
Try to prove the following:
Let $V$ and $W$ be vector spaces over a field $\Bbb F$; let $T$ be a map from $V$ to $W$. 


*

*The following are equivalent:


*

*$T$ is injective.

*$T$ takes each linearly independent set in $V$ to a linear independent set in $W$.


*The following are equivalent:


*

*$T$ is surjective. 

*$T$ takes each spanning set of $V$ to a spanning set of $W$.


