Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $B$?

share|improve this question
1  
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
1  
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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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