Let A be $3\times3$ matrix such that $Au$ and $u$ are orthogonal for each column vector $u\in\Bbb R^3$. Prove that $A^T=-A$. I thought $(A+A^T)u\cdot u=0$ but it doesn't imply $A^T=-A$, so I'm stuck...
Edit : Does this property also hold when we replace 3 by an arbitrary $n$?
We have for all $u \in \mathbb{R}^n$,
$(Au)^Tu = 0$
setting $u = e_i$, we see that $a_{ii} = 0$ $\forall i$
setting $u = e_i + e_j$, we have
$(Au)^Tu = 0$
$\iff (Ae_i + Ae_j)^T(e_i + e_j) = 0$
$\iff (a_i^T + a_j^T)(e_i + e_j) = 0$
$\iff a_i^Te_i + a_j^Te_i + a_i^Te_j + a_j^Te_j = 0$
$\iff a_{ij} + a_{ji} = 0$
where $a_i$ denotes the $i^{th}$ column of $A$
this shows $A + A^T = 0$
I'll assume the size $n$ of the square matrices is arbitrary. To a square matrix $A$ is associated a quadratic form $\Bbb R^n\to\Bbb R$ given by $v\mapsto v\cdot Av$; the question is to show that if this form is identically zero, then $A$ is skew-symmetric. In fact the implication goes both ways.
Every real square matrix is uniquely the sum of a symmetric and a skew-symmetric matrix, and the associated quadratic form then is the sum of those associated to the symmetric and a skew-symmetric summands. We must show that this quadratic form is identically zero if and only if the symmetric part is zero.
So first assume the symmetric part is zero, so we have an skew-symmetric matrix $A$. Then for all $v\in\Bbb R^n$ one has $v\cdot Av=A^Tv\cdot v=v\cdot(-A)v=-(v\cdot Av)$, from which we solve $v\cdot Av=0$ so indeed the quadratic form is identically $0$.
Since we now know that the skew-symmetric summand does not contribute to the quadratic form, it remains to show that the quadratic form $Q$ for a nonzero symmetric matrix $A$ is not identically zero. One could argue from the spectral theorem that $A$, being diagonalisable, must have a nonzero eigenvalue, and a corresponding eigenvector$~v$ gives an example of a vector with $Q(v)\neq0$. But one can also do without invoking the spectral theorem. The symmetric bilinear form $\varphi:\Bbb R^n\times\Bbb R^n\to\Bbb R$ given by $\varphi(v,w)=v\cdot Aw$ associated to the matrix is obviously not identically zero (for instance take $w\notin\ker(A)$ and $v=Aw$), and we can show that $\varphi$ is determined by$~Q$, so that $Q=0$ would imply $\varphi=0$, which is not the case. The expression of $\varphi$ in terms of $Q$ is done by a so-called polarisation formula; there are (at least) two of them to choose from: $$ \varphi(v,w)=\frac14\bigl(Q(v+w)-Q(v-w)\bigr) $$ and $$ \varphi(v,w)=\frac12\bigl(Q(v+w)-Q(v)-Q(w)\bigr); $$ both can be proved using $Q(v)=\varphi(v,v)$ and bilinearity and symmetry of $\varphi$.
Expand out $$ \pmatrix{u_1 & u_2 & u_3} \pmatrix{a_{11} & a_{12} & a_{13}\cr a_{21} & a_{22} & a_{23}\cr a_{31} & a_{32} & a_{33}} \pmatrix{u_1\cr u_2\cr u_3} = 0$$ The coefficients of each monomial $u_1^2, u_1u_2, \ldots, u_3^2$ all must be $0$. These should tell you exactly that $A^T = -A$.
