Suppose the set were not linearly independent so there are not all zero s.t. . Apply T to this equation to reach a contradiction.
For the second, try showing that .
Let be a linear transformation.
If are vectors in such that the set is linearly independent in , prove that the set is linearly independent.
If and, for every linearly independent set in , the set is always linearly independent, prove that there is an inverse transformation . (i.e. are both the identity transformation).
This one is pretty tricky, needless to say.
Kelvin function? I meant the kernel lol.
1) If then meaning . But the are linearly independent so the are identically zero.
2) Let be a basis for the vector space. Let be a vector in the kernel of T. Then . But because are linearly independent, this implies that the 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!
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 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 and suppose that Tx=0. Then so by linear independence of the , the 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.