A scalar valued function is defined as f(X) = XTAX + bTX + c, where A is a symmetric positive definite matrix with dimension n × n; b and x are vectors of dimension n × 1. The minimum value of f(X) will occur when X equals A. (ATA)-1b B. -(ATA)-1b C. \[ – \left( {\frac{{{{\text{A}}^{ – 1}}{\text{b}}}}{2}} \right)\] D. \[\frac{{{{\text{A}}^{ – 1}}{\text{b}}}}{2}\]

” option4=”\[\frac{{{{\text{A}}^{ – 1}}{\text{b}}}}{2}\]” correct=”option4″]

The correct answer is $\boxed{\frac{{{{\text{A}}^{ – 1}}{\text{b}}}}{2}}$.

Let $X$ be a vector of dimension $n \times 1$. The function $f(X)$ is defined as follows:

$$f(X) = XTAX + bTX + c$$

where $A$ is a symmetric positive definite matrix with dimension $n \times n$; $b$ and $x$ are vectors of dimension $n \times 1$.

We can write the function $f(X)$ as follows:

$$f(X) = X^T(A^T + A)X + b^TX + c$$

Since $A$ is a symmetric positive definite matrix, it is invertible. Therefore, we can write the function $f(X)$ as follows:

$$f(X) = (X^TA^TX + b^TX + c)(A^{-1})^T$$

We can minimize the function $f(X)$ by setting its derivative to zero. The derivative of $f(X)$ with respect to $X$ is given by:

$$\frac{\partial f(X)}{\partial X} = 2(A^T + A)X + b^T = 2AX + b^T$$

Setting the derivative to zero and solving for $X$, we get:

$$X = -(A^T + A)^{-1}b^T = -A^{-1}b^T$$

Therefore, the minimum value of $f(X)$ will occur when $X = -A^{-1}b^T$.

The other options are incorrect because they do not minimize the function $f(X)$.