If a quadratic form is represented as a matrix, it may not be symmetric, but there must exist a representation that is symmetric (or Hermitian if complex). But is the reverse true? That is, do all symmetric and Hermitian matrices represent a quadratic form (except maybe in the case where matrices are all zeros)?
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2For a symmetric matrix $A$, consider the map $x\mapsto x^T Ax$ (in the real case). What are the properties that it needs to satisfy to be a quadratic form? Can you check them? – ViktorStein Nov 15 '23 at 19:10
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$A$ represents a quadriatic form if $x^TAx = x^TBx$ where $B$ is unique and where $B=\frac{A +A^T}{2}$ – Davey Nov 15 '23 at 19:33
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I'm not sure what you mean exactly but note that since $A$ is symmetric we have $A + A^T = 2A$. – ViktorStein Nov 16 '23 at 08:32
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1What you wrote is true. Any quadratic form can be represented by a symmetric matrix. The reverse is trivially true. The form that is identically zero is a quadratic form. Perhaps some prefer to call it degenerate. – Kurt G. Nov 16 '23 at 08:48
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Roughly speaking, for a field $F$ with $char F\ne 2$, there is a correspondence between symmetric bilinear forms and quadratic forms (Proposition 1.6 in Pfister's Book). So, the answer for your question is "yes" for real/complex symmetric matrices.
For a more detailed exposition on quadratic forms over fields (and a proof of the above cited Proposition) I recommend Introduction To Quadratic Forms Over Fields or Quadratic Forms with Applications to Algebraic Geometry and Topology.
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Ok so, for char ≠ 2
- every matrix has an associated bilinear form
- every symmetric matrix has an associated bilinear form
- every symmetric bilinear form has an associated quadratic form