Can I deduce that if $\displaystyle \sum_{i \neq j}^{n} x_i x_j a_{ij} = 0 \implies a_{ij}=0?$ Just so you don’t get confused, this could also be written as $\displaystyle \sum_{i=1}^{n}\sum_{j=1}^{n} x_i a_{ij}x_j \implies a_{ij}=0.$ Note: $A$ is a symmetric matrix and $a_{ij} $are its entries. Also this is related to calculating $x^TAx$ for a symmetric matrix $A$. Note 2: The original question, though, is an assignment question which asks to prove that for a symmetric matrix $A \in M_{n\times n}(\mathbb{R})$ if $x^TAx=0$ for all $x \in \mathbb{R}^n$ then $A=0_{n\times n}.$ I would appreciate if you don’t reveal too much of the solution to the assignment question if possible, since I’d rather like to solve it myself. Instead focus on the smaller statement I have stated above. If you think that the deduction is perfectly logical, then please explain the method. I seem to have worked it out for $n=3$ or similar small values but a calculation for arbitrary n seems messy enough that I feel that there must be an easier way to do that.
2026-04-03 06:22:16.1775197336
Linear Algebra Quadratic Form
349 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail AtRelated Questions in LINEAR-ALGEBRA
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