Let $V$ be a finite dimensional inner product space, $ \phi :V \rightarrow V$ a linear operator and $\phi^*:V \rightarrow V $ its adjoint.

I wish to show:

$\phi^*$ is an inverse to $\phi$ if and only if $ \langle \phi (v), \phi (w) \rangle = \langle v,w\rangle \ \ \forall \ v,w \in V $

I understand that $\phi^* \circ\ \phi = id_V $ if and only if $ \langle \phi (v), \phi (w) \rangle = \langle v,w\rangle \ \ \forall \ v,w \in V. $

The proof I'm reading then states, by rank-nullity, it is easily deduced that $\phi \circ\ \phi^* = id_v $.

How do I show the last part?

Here are the definitions I'm working with:

An inner-product is positive definite, linear in the second variable and conjugate symmetric.

An adjoint, $\phi^*$, is a linear operator with the property: $ \forall v,w \in V, \langle \phi^*(v), w \rangle = \langle v, \phi(w) \rangle $


1 Answer 1


Note that it is generally true that for a finite dimensional space $V$ and linear operators $\psi,\psi':V \to V$, we have $$ \psi\circ \psi' = \operatorname{id}_V \implies \psi' \circ \psi = \operatorname{id}_V $$ This is sufficient to complete your proof.

  • $\begingroup$ Thanks :) Is there a name for this proposition? $\endgroup$
    – Bysshed
    Nov 24, 2014 at 20:59
  • $\begingroup$ No name that I'm aware of. An algebraic way to put it is to say that "the ring $\Bbb F^{n \times n}$ is Dedekind finite". $\endgroup$ Nov 24, 2014 at 21:10

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