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Let $M_{n\times n}(R)$ be the set of all $n \times n$ square matrices with real entries, and S the set of all symmetric invertible matrices in $M_{n\times n}(R)$.

For every $P$ in $S$, is there a basis $B$ of an n-dimensional inner product space $V$ with an inner product < , > such that $P$ is the matrix representation of the inner product with respect to the basis $B$?

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No. For example, the $1 \times 1$ matrix $M=-1$ is not the matrix representation of any inner product. –  alex Sep 27 '11 at 2:33
But it is the representation of the bilinear form $-xy$. As i mention in my answer, some authors talk about inner products in a more general sense. –  Manos Sep 27 '11 at 2:40

1 Answer 1

Yes. However, we need to be careful in what we mean by "inner product". If $P$ is only symmetric, then it represents a bilinear form, which some authors refer to as inner product (e.g. Steven Roman). To get positivity, i.e. $<x,x>$ positive for non-zero $x$, we need $P$ to be additionally positive-definite.

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