For each matrix K, vector 𝐟, and scalar c, write out the quadratic function p(𝐱) given by (5.10)

step 1 So in this program, we know that for part A, we know that n less equal to T less or equals to ca

For each matrix K, vector 𝐟, and scalar c, write out the...

For each matrix $K$, vector $\mathbf{f}$, and scalar $c$, write out the quadratic function $p(\mathbf{x})$ given by (5.10). Then either find the minimizer $\mathbf{x}^{\star}$ and minimum value $p\left(\mathbf{x}^{\star}\right)$, or explain why there is none. (a) $K=\left(\begin{array}{rr}4 & -12 \\ -12 & 45\end{array}\right), \mathbf{f}=\left(\begin{array}{c}-\frac{1}{2} \\ 2\end{array}\right), c=3$; (b) $K=\left(\begin{array}{ll}3 & 2 \\ 2 & 1\end{array}\right), \mathbf{f}=\left(\begin{array}{l}4 \\ 1\end{array}\right)$, $$ \begin{aligned} & c=0 ; \quad(c) K=\left(\begin{array}{rrr} 3 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 3 \end{array}\right), \mathbf{f}=\left(\begin{array}{r} 1 \\ 0 \\ -2 \end{array}\right), c=-3 ;(d) K=\left(\begin{array}{rrr} 1 & 1 & 1 \\ 1 & 2 & -1 \\ 1 & -1 & 1 \end{array}\right), \\ & \mathbf{f}=\left(\begin{array}{r} -3 \\ -1 \\ 2 \end{array}\right), c=1 ;(e) K=\left(\begin{array}{llll} 1 & 1 & 0 & 0 \\ 1 & 2 & 1 & 0 \\ 0 & 1 & 3 & 1 \\ 0 & 0 & 1 & 4 \end{array}\right), \mathbf{f}=\left(\begin{array}{r} -1 \\ 2 \\ -3 \\ 4 \end{array}\right), c=0 . \end{aligned} $$ (c)

For each matrix $K$, vector $\mathbf{f}$, and scalar $c$, write out the quadratic function $p(\mathbf{x})$ given by (5.10). Then either find the minimizer $\mathbf{x}^{\star}$ and minimum value $p\left(\mathbf{x}^{\star}\right)$, or explain why there is none. (a) $K=\left(\begin{array}{rr}4 & -12 \\ -12 & 45\end{array}\right), \mathbf{f}=\left(\begin{array}{c}-\frac{1}{2} \\ 2\end{array}\right), c=3$;
(b) $K=\left(\begin{array}{ll}3 & 2 \\ 2 & 1\end{array}\right), \mathbf{f}=\left(\begin{array}{l}4 \\ 1\end{array}\right)$,
$$
\begin{aligned}
& c=0 ; \quad(c) K=\left(\begin{array}{rrr}
3 & -1 & 1 \\
-1 & 2 & -1 \\
1 & -1 & 3
\end{array}\right), \mathbf{f}=\left(\begin{array}{r}
1 \\
0 \\
-2
\end{array}\right), c=-3 ;(d) K=\left(\begin{array}{rrr}
1 & 1 & 1 \\
1 & 2 & -1 \\
1 & -1 & 1
\end{array}\right), \\
& \mathbf{f}=\left(\begin{array}{r}
-3 \\
-1 \\
2
\end{array}\right), c=1 ;(e) K=\left(\begin{array}{llll}
1 & 1 & 0 & 0 \\
1 & 2 & 1 & 0 \\
0 & 1 & 3 & 1 \\
0 & 0 & 1 & 4
\end{array}\right), \mathbf{f}=\left(\begin{array}{r}
-1 \\
2 \\
-3 \\
4
\end{array}\right), c=0 .
\end{aligned}
$$
(c)

Key Concepts

Quadratic Function in Multiple Variables

A quadratic function in multiple variables is typically expressed in the form p(x) = x?Kx + 2f?x + c (or a similar variant), where K is a symmetric matrix, f is a vector, and c is a scalar constant. This formulation captures both the quadratic (second-order) behavior through the matrix K and the linear effects via the vector f. Understanding this structure is crucial because it allows one to analyze the function's curvature and determine whether optimization methods can guarantee a unique minimum.

Matrix Definiteness

The definiteness of the matrix K plays a key role in the analysis of quadratic functions. When K is positive definite (i.e., x?Kx > 0 for all nonzero x), the quadratic function is strictly convex and has a unique minimizer. If the matrix is only positive semidefinite, the function may have a set of minimizers, while if K is indefinite or not positive semidefinite, the function may not have a minimum at all, as it could be unbounded below or exhibit saddle point behavior.

Optimization of Quadratic Functions

Optimizing quadratic functions typically involves setting the gradient of the function equal to zero and solving for the variable vector x. Since the gradient of a quadratic function is linear in x (given by ?p(x) = 2Kx + 2f), the condition for optimality leads to a system of linear equations. The solution to this system, provided that K is invertible (particularly in the positive definite case), yields the unique minimizer of the quadratic function.

Completing the Square

Completing the square is a technique that reexpresses the quadratic function in a form that clearly shows the minimizer and minimum value. By grouping and manipulating the quadratic and linear terms, the function can be written as a perfect square plus a constant. This reformulated expression makes it easy to identify the variable values that nullify the quadratic term and hence minimize the function, while also revealing how the constant term shifts the overall minimum value.

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