# Question about Bilinear form and inner product space

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.

## 1 Answer

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. $( \cdot | \cdot )$.