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In Exercises $3-6,$ find (a) the maximum value of $Q(x)$ subject to the constraint $x^{T} x=1,(b)$ a unit vector $u$ where this maximum is attained, and $(c)$ the maximum of $Q(x)$ subject to the constraints $\mathbf{x}^{T} \mathbf{x}=1$ and $\mathbf{x}^{T} \mathbf{u}=0$$$\begin{array}{l}{Q(\mathbf{x})=5 x_{1}^{2}+6 x_{2}^{2}+7 x_{3}^{2}+4 x_{1} x_{2}-4 x_{2} x_{3}} \\ {\text { (See Exercise } 1 . )}\end{array}$$

a) Maximum value of $x^{T} A x$ subject to constraint $x^{T} x=1$ is the greatest eigenvalue $\lambda_{1}$ of $A(\text { see Theorem } 6)$From exercise $1,$ we know that $\lambda_{1}=9$

b) Maximum value of $x^{T} A x$ subject to the constraint $x^{T} x=1$ occurs at a unit eigen vector $u$ corresponding to the greatest eigenvalue $\lambda_{1}$ of $A .$ (again, by Theorem 6 )By exercise $1, u=\pm\left[\begin{array}{c}{1 / 3} \\ {2 / 3} \\ {-2 / 3}\end{array}\right]$

c) Maximum value of $x^{T} A x$ subject to the constraints $x^{T} x=1$ and $x^{T} u=0$$\left.\text { is the second greatest eigenvalue } \lambda_{2} \text { of } A . \text { (see Theorem } 7\right)$By exercise $1, \lambda_{2}=6$

Algebra

Chapter 7

Symmetric Matrices and Quadratic Forms

Section 3

Constrained Optimization

Introduction to Matrices

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so by excess one, Um, the maximum value off this quadratic a phone Q X, um, subject to the constraint, um, X transpose X equals to one Lemonis X is a unit vector. I will be the makes him all Eigen value. So in this case, the Mexi Magan Value is loved. Ah wai In close to nine, that means this is the next Mama Leo off q X subject Teoh any you meet back to Rex and ah one q X equals tonight. So when q x a chief's its maximum value, the corresponding that unit vector will be the item backed her, um, associates to this aggravator of nine. So it's 1/3 um, 12 miners toe and the CCT since this is ah quadratic terms. So, um, the choice off this Eigen vector could be plows miners. So we have two different choices. No, If we add another constraint, that's, um, X transpose times X equals zero. Say so you need vector, um, and the X transpose times you it causes your That means X is perpendicular to on this unions factor. Then the maximum value off this chaotic time, Q X. We will be the second largest agon. Valeo now belong that we caused to six

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