Question

\(A\) is an \(m\times m\) skew-symmetric matrix with real-valued entries, and \(x\) is an \(m\)-dimensional column vector with real-valued entries such that \(x^{T}x=1\). The quantity \(x^{T}Ax\) evaluates to (answer in integer).

Hint:

For any skew-symmetric A, the quadratic form x transpose A x always equals its own negative, so it must be zero.

Answer 1

Correct Answer: 0

Step 1: Write the quadratic form as a double sum.
For any matrix $A$ with entries $a_{ij}$, we can expand $x^{T}Ax=\sum_{i=1}^{m}\sum_{j=1}^{m}a_{ij}x_ix_j$.

Step 2: Use the skew-symmetric property on the entries.
Since $A$ is skew-symmetric, $a_{ji}=-a_{ij}$, and every diagonal entry must satisfy $a_{ii}=-a_{ii}$, which forces $a_{ii}=0$ for every $i$.

Step 3: Pair up the off-diagonal terms.
Group the sum into diagonal terms (all zero) and pairs $(i,j)$ with $i\neq j$. Each pair contributes
\[ a_{ij}x_ix_j+a_{ji}x_jx_i = a_{ij}x_ix_j - a_{ij}x_ix_j = 0 \]
because $x_ix_j=x_jx_i$ but $a_{ji}=-a_{ij}$.

Step 4: Add everything up.
Every diagonal term is zero and every off-diagonal pair cancels, so the whole double sum is zero regardless of what $x$ is.

Step 5: Note on the given condition.
The unit-length condition $x^{T}x=1$ is extra information here; it plays no role since the cancellation happens term by term.
\[ \boxed{0} \]