Let $f: V \to V$ be a linear operator on the finite-dimensional real vector space V. Let B be a basis for V. Let A be the real matrix which represents $f$ with respect to B.
(a) $f$ is invertible if and only if $f(B)$ is a basis for $V$.
So suppose $f$ is invertible. By definition of invertible, then there exists some matrix $C$ such that $f(B)C = Cf(B) = I$. I don't really know where to go from here… I suppose that I need to show that $f(B)$ spans $V$ and linear independence. I don't really know the notation for how to do this or how to draw this conclusion from the fact that $f$ is invertible.
(b) $f$ is invertible if and only if $A$ is an invertible matrix.
By definition $A$ represents $f$ with respect to $B$ and it's a given the $f$ is invertible so obviously $A$ should be to. I don't get what I'm trying to prove here… or how to do it.
I'm having a lot of difficulty with this as my text is very ambiguous with the definitions relating to linear operators. Any help is much appreciated.
Best Answer
HINTS:
(1) Can you show that an invertible linear map sends linearly independent vectors to linearly independent vectors? Then dimension will do the rest for you.
(2) If $f$ is invertible and $A$ is the matrix representing $f$ with respect to $B$, can you guess what matrix might be its inverse?