1. Jun 7, 2014

### mahler1

I have doubts regarding a statement related to the following proposition: Let $(V,<,>)$ be a finite dimensional vector space equipped with an inner product and let $f:V \to V$ be a linear transformation, then the following statements are equivalent:

1)$<f(v),f(w)>=<v,w>$ for all $v,w \in V$
2) $f^* \circ f=f \circ f^*=id_V$

I've read and understood the proof of the equivalence between these two statements but I have a major doubt with the following: In the part 1) $\implies$ 2), in the textbook they prove 1) $\implies f^* \circ f=id_V$ and then affirm "As V is a finite dimensional vector space, $f^* \circ f=id_V \implies f \circ f^*=id_V$. I don't understand why that implication is true, I would appreciate if someone could explain it to me.

2. Jun 8, 2014

### hilbert2

It's proved here in theorem 4, part 7), that in a finite-dimensional vector space the left and right inverses of a transformation are equal: http://www-math.mit.edu/~dav/onesidedCORR.pdf [Broken] .

Last edited by a moderator: May 6, 2017
3. Jun 8, 2014

### Fredrik

Staff Emeritus
The equality $f^*\circ f=\mathrm{id}_V$ implies that f* is surjective. A linear operator on a finite-dimensional vector space is surjective if and only if it's injective. (Can you prove those two statements?) So f* is bijective, and therefore invertible.
\begin{align}
&f^*\circ f =\mathrm{id}_V\\
&\forall x~~ f^*(f(x)) =x\\
&\forall x~~ (f^*)^{-1}(f^*(f(x)))= (f^*)^{-1}(x)\\
&\forall x~~ f(x) = (f^*)^{-1}(x)\\
&f =(f^*)^{-1}(x)
\end{align} You could also just "apply $(f^*)^{-1}$ from the left" to both sides of the first equality, and then conclude that $f=(f^*)^{-1}$, but that would be a bit suspicious IMO, because you would be using that $\circ$ is associative, and maybe what you have proved earlier is just that $(f\circ g)\circ h=f\circ (g\circ h)$ when f,g,h are all bijective. (We don't know that f is bijective at the start of the calculation. That's a conclusion we can make when we get to line 4).