A is an m×m skew-symmetric matrix with real-valued entries, and x is an m dimensional column vector with real-valued entries such that x^Tx = 1. The quantity x^TAx evaluates to ___________(Answer in integer)

The correct option is 0.

To understand why the quantity evaluates to this value, let us break down the mathematical steps using the properties of transposes and skew-symmetric matrices.

First, let us denote the quantity as a scalar value $y$:

$y=x^{T}Ax$

Since $y$ is a scalar (a 1×1 matrix), taking its transpose does not change its value. Therefore, we can write:

$y=y^T$

Substituting the expression for $y$ into this relation, we get:

$x^{T}Ax={\left(x^{T}Ax\right)}^T$

Using the matrix transpose property ${\left(PQR\right)}^T=R^T{Q}^T{P}^T$, we expand the right side of the equation:

${\left(x^{T}Ax\right)}^T=x^T{A}^T{\left(x^T\right)}^T$

Since the double transpose of a vector yields the original vector, ${\left(x^T\right)}^T=x$, the expression simplifies to:

${\left(x^{T}Ax\right)}^T=x^T{A}^{T}x$

Now, we use the definition of a skew-symmetric matrix. By definition, a matrix $A$ is skew-symmetric if its transpose is equal to its negative:

$A^T=-A$

Substituting $A^T=-A$ into our transposed expression gives:

$x^T{A}^{T}x=x^T(-A)x=-x^{T}Ax$

Combining the equations, we have established that:

$x^{T}Ax=-x^{T}Ax$

Adding $x^{T}Ax$ to both sides of the equation yields:

$2\left(x^{T}Ax\right)=0$

Dividing by 2, we conclude:

$x^{T}Ax=0$

This result holds for any skew-symmetric matrix $A$ and any real vector $x$, regardless of the vector's length constraint.