1. Linear transformation proof problem

Let $T:R^n\rightarrow R^m$ be a linear transformation.
If $v_1,v_2,...,v_k$ are vectors in $R^n$ such that the set $\{T(v_1),T(v_2),...,T(v_k)\}$ is linearly independent in $R^m$, prove that the set $\{v_1,v_2,...,v_k\}$ is linearly independent.
If $n=m$ and, for every linearly independent set $\{w_1,w_2,...,w_k\}$ in $R^n$, the set $\{T(w_1),T(w_2),...,T(w_k)\}$ is always linearly independent, prove that there is an inverse transformation $S:R^n\rightarrow R^n$. (i.e. $S\circ T, T\circ S$ are both the identity transformation).

This one is pretty tricky, needless to say.

2. Suppose the set $\{v_i\}$ were not linearly independent so there are $a^i$ not all zero s.t. $a^iv_i=0$. Apply T to this equation to reach a contradiction.

For the second, try showing that $\mathrm{ker}\; T = 0$.

Suppose the set $\{v_i\}$ were not linearly independent so there are $a^i$ not all zero s.t. $a^iv_i=0$. Apply T to this equation to reach a contradiction.

For the second, try showing that $\mathrm{ker}\; T = 0$.
We don't have the Kelvin function available, so that wouldn't work.

Also, though I could probably write out all of the first part, I'm tied up with other Algebra-related material, so do you think you could show a bit more detail?

4. Kelvin function? I meant the kernel lol.

1) If $a^iv_i=0$ then $T(a^iv_i)=T(0)$ meaning $a^iT(v_i)=0$. But the $T(v_i)$ are linearly independent so the $a^i$ are identically zero.

2) Let $v_i$ be a basis for the vector space. Let $x=x^iv_i$ be a vector in the kernel of T. Then $T(x)=x^iT(v_i)=0$. But because $T(v_i)$ are linearly independent, this implies that the $x^i$ are identically zero so x=0. So the kernel of T is zero. It follows that T is injective. Now show it is surjective and use these to deduce it is invertible.

P.S. I use the Einstein summation convention. I'm going to get in trouble if I don't get back to work now lol. Good luck!

Kelvin function? I meant the kernel lol.

1) If $a^iv_i=0$ then $T(a^iv_i)=T(0)$ meaning $a^iT(v_i)=0$. But the $T(v_i)$ are linearly independent so the $a^i$ are identically zero.

2) Let $v_i$ be a basis for the vector space. Let $x=x^iv_i$ be a vector in the kernel of T. Then $T(x)=x^iT(v_i)=0$. But because $T(v_i)$ are linearly independent, this implies that the $x^i$ are identically zero so x=0. So the kernel of T is zero. It follows that T is injective. Now show it is surjective and use these to deduce it is invertible.

P.S. I use the Einstein summation convention. I'm going to get in trouble if I don't get back to work now lol. Good luck!
Thanks for your help on the first half, but as for the second half, we don't have kernels available. I got the name mixed up last post.

6. There is no reason not to use kernels. I'm not sure what you mean by them not being available..... The kernel of a map is the set of vectors it sends to zero. There is a beautiful relationship between the kernel and the image of a map called the rank-nullity theorem.

...but in any case, this particular proof doesn't use them in any essential way so I'll reformulate it without them:

Lemma. If T is linear and $Tv=0\implies v=0$ then T is injective.

Proof. For suppose not. Then there are distinct x and y s.t. Tx=Ty. But then T(x-y)=0, which implies that x-y=0, or x=y, contrary to our assumption.

Then to prove your 2) Take an $x=x^iv_i$ and suppose that Tx=0. Then $Tx=x^iTv_i=0$ so by linear independence of the $Tv_i$, the $x^i$ are all zero, meaning x=0. Therefore if Tx=0, then x=0, and we may apply the lemma to see that T is injective. Surjectiveness is easy and so T is invertible.