0
$\begingroup$

This is a question I have stumbled upon in a test I found on the web, and I don't even know how to approach it:

Say $V$ is a vector space with an inner product above $\mathbb{C}$ (We don't know what is the inner product). Prove that if a linear operator $f\colon V\to V$ is adjoint to itself, the bilinear form $B\colon V\times V\to \mathbb{C}$ defined by $B(v,w)=(f(v)\mid w)$ is a Hermitian form.

Now, I understood that an adjoint operator $F$ is an operator that fulfills: $$\langle F(v),u\rangle = \langle v,F(u)\rangle$$ Now, what is $(f(v)\mid w)$ and what is the definition of an adjoint bilinear form? I only found the definition of an adjoint operator so that's why i'm asking.

Help would be very appreciated.

$\endgroup$
0
$\begingroup$

The bilinear form should be defined as $$ B(v,w) = \langle f(v), w \rangle, \ v,w\in V, $$ note that it uses the inner product! Now use the definition of self-adjointness $$ \langle f(v), w \rangle = \langle v, f(w) \rangle \quad\forall v,w\in V $$ to conclude $$ B(v,w) = \overline{ B(w,v) } \quad\forall v,w\in V. $$ To me it looks like the confusion comes from different notations for inner product: $\langle \cdot,\cdot \rangle$ vs. $( \cdot | \cdot )$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.