Symmetric matrix result: $A^{T} + A = S$

Question

I came across this question in a paper, there are no solutions.

Let $S \in \text{Mat}_{n\times n}(\mathbb{R})$ be a symmetric matrix, having the property that $v^{t}Sv > 0$, $v\in \mathbb{R}^n \setminus \{ \mathbf{0} \}$.

Suppose $A \in \text{Mat}_{n\times n}(\mathbb{R})$ s.t. $A^{T} + A = S$.

Prove that the null space of $A$ is $\{ \mathbf{0} \}$.

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I think of a few ways to do this, but I'm not sure. For example we can use a $3\times3$ matrix:

$$ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} + \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}^T = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} + \begin{pmatrix} a & d & g \\ b & e & h \\ c & f & i \end{pmatrix} = \begin{pmatrix} 2a & b+d & c+g \\ b+d & 2e & f+h \\ c+g & f+h & 2i \end{pmatrix} $$

However, this does not look very elegant. Is there a result regarding symmetric matrices I'm missing?

Answer 1

If $v \neq 0, \;Av = 0$ then $$v^T Sv = v^T(A + A^T)v = v^TAv + v^TA^Tv = 0 + 0=0$$

Which contradicts your condition (which is called positive definiteness).

Answer 2

Hint: Assume the null space of $A$ is not $\left\{0\right\}$. Therefore there exists $v\in\mathbb{R}^{n}$ such that $Av=0$ and as a consequence $v^{t}A^{t}=0$ as well. Can you calculate $v^{t}Sv$?