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: Recall what skew-symmetric means.
A matrix \(A\) is skew-symmetric when its transpose equals its own negative:
\[ A^{T}=-A \]

Step 2: Notice that \(x^{T}Ax\) is just a number.
Since \(x\) is an \(m\times1\) column vector and \(A\) is \(m\times m\), the product \(x^{T}Ax\) is a \(1\times1\) matrix, in other words a plain scalar. A scalar always equals its own transpose.

Step 3: Take the transpose of the scalar.
\[ (x^{T}Ax)^{T}=x^{T}A^{T}x \]
Since the left side is a scalar, it equals the right side, so
\[ x^{T}Ax=x^{T}A^{T}x \]

Step 4: Substitute the skew-symmetric property.
Using \(A^{T}=-A\),
\[ x^{T}Ax=x^{T}(-A)x=-x^{T}Ax \]

Step 5: Solve for the value.
\[ x^{T}Ax=-x^{T}Ax\ \Rightarrow\ 2x^{T}Ax=0\ \Rightarrow\ x^{T}Ax=0 \]

Step 6: Note about the condition \(x^{T}x=1\).
This condition just says \(x\) has unit length. It does not change the result above; the quantity \(x^{T}Ax\) is zero for any real vector \(x\), not only for unit vectors.

Final Answer:
\[ \boxed{0} \]