ThmDex – An index of mathematical definitions, results, and conjectures.
Real square matrix antisymmetric part is zero definite
Formulation 0
Let $A \in \mathbb{R}^{N \times N}$ and $x \in \mathbb{R}^{N \times 1}$ each be a D4571: Real matrix such that
(i) $N \in 1, 2, 3, \ldots$ is a D5094: Positive integer
Then \begin{equation} x^T \left( \frac{A - A^T}{2} \right) x = 0 \end{equation}
Proofs
Proof 0
Let $A \in \mathbb{R}^{N \times N}$ and $x \in \mathbb{R}^{N \times 1}$ each be a D4571: Real matrix such that
(i) $N \in 1, 2, 3, \ldots$ is a D5094: Positive integer
Since $x^T A x \in \mathbb{R}^{1 \times 1}$, then $x^T A x$ is symmetric due to R3987: 1-by-1 matrices are always symmetric. That is, $(x^T A x)^T = x^T A x$. Applying R3747: Transpose of finite product of real matrices, we then have \begin{equation} x^T A^T x = (x^T A x)^T = x^T A x \end{equation} Subtracting $x^T A^T x $ from each side, we get $x^T A x - x^T A^T x = 0$. Thus \begin{equation} x^T \left( \frac{A - A^T}{2} \right) x = \frac{1}{2} \left( x^T A x - x^T A^T x \right) = \frac{1}{2} \cdot 0 = 0 \end{equation} $\square$