TITLE: Prove that sum of matrices equals zero
QUESTION [1 upvotes]: The matrix $A$ has size $3 \times 3$ and we know that for any column
  vector $v\in \mathbb{R}^{3}$ the vectors $Av$ and $v$ are orthogonal. Prove that
  $A^{T} + A = 0$, where $A^{T}$ is the transposed matrix $A$.

So if 
$$A = \begin{pmatrix}
    a_{11}  & a_{12} & a_{13}\\
    a_{21}  & a_{22} & a_{23}\\
    a_{31}  & a_{32} & a_{33}
\end{pmatrix}$$
and
$$v = \begin{pmatrix}
    x\\
    y\\
    z  
\end{pmatrix}$$
orthogonality of $Av$ and $v$ brought me to the equation
$x \cdot y \cdot (a_{12}+a_{21}) + x \cdot z \cdot (a_{13}+a_{31}) + y \cdot z \cdot (a_{23}+a_{32}) + a_{11} \cdot x^{2} + a_{22} \cdot y^{22} + a_{33} \cdot z^{2} = 0$
and the matrix $A^{T} + A$ equals
$$A^{T} + A = \begin{pmatrix}
    2a_{11}  & a_{12} + a_{21} & a_{13} + a_{31}\\
    a_{12} + a_{21}  & 2a_{22} & a_{23} + a_{32}\\
    a_{13} + a_{31} & a_{23} + a_{32} & 2a_{33}
\end{pmatrix}$$
However I don't see how then prove that $A^{T} + A = 0$.

REPLY [1 votes]: Following your working,
$x \cdot y \cdot (a_{12}+a_{21}) + x \cdot z \cdot (a_{13}+a_{31}) + y \cdot z \cdot (a_{23}+a_{32}) + a_{11} \cdot x^{2} + a_{22} \cdot y^{22} + a_{33} \cdot z^{2} = 0 \tag{1}$
Note that equation $(1)$ holds for any choice of $v=\begin{pmatrix}x \\ y \\ z \end{pmatrix}$. That is we can choose $x, y, z$.
If you substitute $x=1, y=0, z=0$, we conclude that $a_{11}=0$.
If you substitute $x=0, y=1, z=0$, we conclude that $a_{22}=0$.
If you substitute $x=0, y = 0, z=1$, we conclude that $a_{33}=0$.
By now, we can conclude that the diagonal entries of $A+A^T$ are all zeros.
Hence equation $(1)$ now reduces to 
$x \cdot y \cdot (a_{12}+a_{21}) + x \cdot z \cdot (a_{13}+a_{31}) + y \cdot z \cdot (a_{23}+a_{32})  = 0 \tag{2}$
If we substitute $x=1,y=1, z=0$, we conclude that $a_{12}+a_{21}=0$. 
If we substitute $x=1, y=0, z=1$, we conclude that $a_{12}+a_{31}=0$.
If we substitue $x=0, y=1, z=1$, we conclude that $a_{23}+a_{32}=0$
Hence we can conclude that the off diagonal of $A+A^T$ are zeros as well. 
Hence we have $A+A^T=0$.

The result actually hold for square matrices $A$ of general size $n \times n$. That is if $A \in \mathbb{R}^{n \times n}$ and for any $v \in \mathbb{R}^n$, if $v$ and $Av$ are orthogonal, then $A+A^T=0$.
From the condition that $v$ and $Av$ are orthogonal, we have $v^TAv=0$.
We let $e_i$ be the standard unit basis with the $i$-th position being $1$ and the other positions being $0$. For example in the case of $n=3$, we have $e_2 = \begin{pmatrix} 0 \\ 1\\ 0\end{pmatrix}$. These matrices are very useful, we have $$e_i^TAe_j = A_{ij}$$
because you can check that $Ae_j$ would be just the $j$-th column of $A$ and then $e_i$ would select the $i$-th row.
We first handle the diagonal entries. We know that $e_iAe_i=0$, this is just saying that $A_{ii}=0$, hence the diagonal part s of $A+A^T=0$.
Also, we know that \begin{align}(e_i+e_j)^TA(e_i+e_j)&=e_i^TAe_i+e_i^TAe_j+e_j^T Ae_i + e_j^TAe_j \\
&= A_{ii} + A_{ij}+A_{ji}+A_{jj}\\
&= A_{ij}+A_{ji}\end{align}
and we know that it is equal to $0$.
Hence $A+A^T=0$.
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