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.

• $$\sum_{i\neq j}^nx_ix_ja_{ij}\neq\sum_{i=1}^n\sum_{j=1}^nx_ia_{ij}x_j.$$ – José Carlos Santos Oct 21 '17 at 8:06
• you mean $a_{ij}=0$ for $i\neq j$ ??? – Isham Oct 21 '17 at 8:12
• Let $x = e_k$, that is the $k^{th}$ column of the Identity Matrix. Then we get $x^T A x = a_{kk}$. So $a_{kk}=0$. Hence all the entries in the main diagonal are 0 for $A$. What remains, is to check for the other non-diagonal entries. Since $x^T A x = \displaystyle\sum_{i=1}^{n} \displaystyle\sum_{j=1}^{n} x_i a_{ij} x_j = \displaystyle\sum_{i=1}^n x_i^2 a_{ii} + 2 \displaystyle\sum_{i\neq j}^{n} x_i x_j a_{ij}$, $x_{ii}=0$ just makes our lives easier. So the first series sums to $0$. The second series set equal to $0$ gives rise to the expression that I mentioned in the question! – Aalekh Patel Oct 21 '17 at 13:12

You have already worked out that the diagonal elements of $A \ (a_{kk}) = 0$.
\eqalign{ \therefore x^TAx &= \sum_{i=1}^na_{ii}x_i^2 + \sum_{i\neq j}a_{ij} x_i x_j \\ &= \sum_{i\neq j}a_{ij} x_i x_j }
Now consider $x$ such that $x_k = x_l = 1$ and $x_i = 0 \ \forall i\neq k,l$. This reduces the above expression to
\eqalign{ x^TAx &= a_{kl} + a_{lk} }
Equating it with $0$, we get: $a_{kl} = -a_{lk} \ \forall k \neq l$
This proves the more general property that $A^T = -A$ for any square matrix $A$, and in the special case of $A$ being symmetric, $A = 0$.