Let the superscript T represent the transpose operation. Consider the function $f(x)=\frac{1}{2}{x}^{T}Qx-r^{T}x$, where x and r are n × 1 vectors and Q is a symmetric n × n matrix. the stationery point of f(x) is

The correct option/answer is: Q^-1r.

Step-by-Step Explanation:

We are given the quadratic function:
$$f(x)=\frac{1}{2}{x}^{T}Qx-r^{T}x$$
where:
• $x$ and $r$ are column vectors of size $n\times 1$,
• $Q$ is a symmetric matrix of size $n\times n$ (so $Q^T=Q$),
• the superscript T represents the transpose operation.

To find the stationary point of $f(x)$, we need to compute the gradient of the function with respect to the vector $x$ and set it equal to zero.

1. Differentiating the Quadratic Term:
Let us consider the term $g(x)=x^{T}Qx$. The gradient of a quadratic form with respect to $x$ is given by:
$${\nabla }_x(x^{T}Qx)=(Q+Q^T)x$$
Since $Q$ is symmetric ($Q^T=Q$), this simplifies to:
$${\nabla }_x(x^{T}Qx)=2Qx$$

2. Differentiating the Linear Term:
Now consider the linear term $r^{T}x$. The gradient of this term with respect to $x$ is:
$${\nabla }_x(r^{T}x)=r$$

3. Finding the Gradient of the Function:
Combining the gradient terms, we get:
$${\nabla }_{x}f(x)={\nabla }_x[\frac{1}{2}{x}^{T}Qx-r^{T}x]$$
$${\nabla }_{x}f(x)=\frac{1}{2}(2Qx)-r=Qx-r$$

4. Solving for the Stationary Point:
To find the stationary point, we set the gradient to zero:
$$Qx-r=0$$
Rearranging the terms yields:
$$Qx=r$$
Multiplying both sides by the inverse matrix $Q^{-1}$ gives the stationary point:
$$x=Q^{-1}r$$

As verified from the attached image, analyzing the simpler one-dimensional scalar case where $n=1$ confirms the exact same result: the function simplifies to $f(x)=\frac{1}{2}Q{x}^2-rx$. Taking the derivative yields $f^{\prime }(x)=Qx-r=0$, leading directly to $x=\frac{r}{Q}=Q^{-1}r$. Therefore, the generalization to higher dimensions yields the stationary point $Q^{-1}r$.