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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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up vote 1 down vote accepted

Every self-adjoint matrix can be diagonalized in an orthonormal basis and has real eigenvalues, that's the spectral theorem for matrices.

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.

Therefore, a matrix is self-adjoint for some inner product iff it is diagonalizable, and its eigenvalues are all real.

As to your second question, a scalar product is really just a positive-definite sesquilinear form (bilinear in the real case) that is conjugate-symmetric (or just symmetric in the real case), so you really have answered it.

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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
@Haatschii : as I have said, that's the spectral theorem for matrices. A sketch of the proof can be found on the Wikipedia article (for complex spaces), if that's what you're asking about. – tomasz Jun 19 '12 at 16:31
Got it! Seems I've misread your first paragraph. Thanks – Haatschii Jun 19 '12 at 16:49

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