Given linear operator $A: R^3 \to R^3$, $\forall x \in R^3: (Ax, x)=0$, prove that A is skew-symmetric operator.

Question

Given linear operator $A: R^3 \to R^3$, $\forall x \in R^3: (Ax, x)=0$, prove that A is skew-symmetric operator. I came up with a solution like: if $(Ax, x) = 0 \implies (x, A^{T}x)=(A^{T}x, x)=0 \implies (Ax, x) + (A^{T}x, x)= 0 = ((A+A^{T})x,x)=0 \ \forall x$ then $A+A^{T}=0$, but I'm not sure that last implication is correct.

Answer 1

You're going to need to make more direct use of $(A x, x) = 0$. Specifically, consider that the $(i,j)^{\text{th}}$ element of $A$ (call it $a_{i, j}$) can be found by $$a_{i, j} = (A e_j, e_i)$$

We can conclude that $A$ is skew symmetric if we can show $a_{i, j} = - a_{j, i}$ for all $i, j$. To take advantage of the hypothesis, let's try using $x = e_i + e_j$: $$ 0 = (Ax, x) = (A(e_i + e_j), e_i + e_j) = (Ae_i, e_i) + (Ae_i, e_j) + (Ae_j, e_i) + (Ae_j, e_j) $$

Of course, we already know that $(Ae_i, e_i) = (Ae_j, e_j) = 0$ by hypothesis. Thus we have $$ (Ae_i, e_j) + (Ae_j, e_i) = a_{j, i} + a_{i, j} = 0 $$

which is what we wanted to show.