Originally Posted by

**maddas** Kelvin function? I meant the kernel lol.

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

2) Let $\displaystyle v_i$ be a basis for the vector space. Let $\displaystyle x=x^iv_i$ be a vector in the kernel of T. Then $\displaystyle T(x)=x^iT(v_i)=0$. But because $\displaystyle T(v_i)$ are linearly independent, this implies that the $\displaystyle x^i$ 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!