# How to find every possible scalar product on $V$

I am given the following task:

"For which $a$, $b \in \mathbb{R}$ there exists a scalar product, such that $$A = \left( \begin{matrix} 0 & 0 & a b \\ 1 & 0 & a \\ 0 & 1 & b\end{matrix}\right)$$ is a self-adjoint matrix".

A hint says, the task breaks down in finding $a$, $b$ such that $A$ is diagonalizable. But I can't figure out why.

As far as I got is, that Self-adjoint means that $\left< A v, w\right> = \left< v, A w\right> \forall v, w \in V$. So we are looking for $a,b$ such that there is a scalar product on $V$ such that this equality holds. I know that for every Bilinearform $f$ there is a Matrix $B$ such that $f(v,w) = <v,Bw>$ in means of the canonical scalar product. Is there a similar way to represent every scalar product in means of the canonical scalar product? And anyone has a hint why this should break down to find $a,b$ such that $A$ is diagonalizable?

Thanks for any help

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On the other hand, the converse is also true (it's easy to check), and for every basis, we can construct an inner product with respect to which it is orthonormal (just by inducing it through a bijection with $\mathbf C^n$ which maps the basis onto the standard basis of $\mathbf C^n$), so in particular, the basis diagonalizing a given matrix can be made orthonormal.
As far as I understand it follows $A$ can be diagonalized $\Rightarrow A$ is self-adjoint with respect to the scalar product that includes the bijection to the basis where $A$ is diagonal. So far so good. But why is this a nessesary condition, meaning why can't there be a non diagonalizable that is self-adjoint with respect to some scalar product? –  Haatschii Jun 19 '12 at 8:38