e642a3…: “Far from the great prime”

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Compute the maximum and minimum values attained by the quadratic form $Q(\vec {x})= \vec {x}^TA\vec {x}$ , where $A = \begin {bmatrix} -2&4\\4&4\end {bmatrix}$ subject to the constraint $\vec {x}\vec {x}^{T} = 1$ , and determine a unit vector where each extremum is attained.

Maximum = is attained at $\vec {u} = \frac {1}{\sqrt {\answer {5}}}\begin {bmatrix} \answer {1} \\ \answer {2}\end {bmatrix}$
Minimum = is attained at $\vec {v}= \frac {1}{\sqrt {\answer {5}}}\begin {bmatrix} \answer {2} \\ \answer {-1} \end {bmatrix}$

Find the maximum value for the quadratic form corresponding to the matrix $A$ and the vector $\vec {u}$ you computed above, subject to $\vec {x}\vec {x}^{T}=1$ and $\vec {x}^{T}\vec {u}=0$ .

Maximum = is attained at $\vec {v}= \frac {1}{\sqrt {\answer {5}}}\begin {bmatrix} \answer {2} \\ \answer {-1} \end {bmatrix}$

Find the maximum value for the quadratic form:
$Q(\vec {x}) = 7x_1^2 +3x_2^2 +8x_1x_2$ , subject to $\| \vec {x}\| =1$ .

Maximum = $\answer {5}+\answer {2}\sqrt {\answer {5}}$ .

Given the orthogonal diagonalization of $A$ below: $A = PDP^{-1} \text { where } P = \begin {bmatrix} -2/3 &-1/3&2/3\\ -2/3&2/3&-1/3\\ 1/3& 2/3& 2/3\end {bmatrix} \text { and } D= \begin {bmatrix} 3&0&0\\0&9&0\\0&0&15\end {bmatrix}$ ,

Find the unit vector $\vec {u}$ at which the quadratic form $Q(\vec {x}) = \vec {x}^{T}A\vec {x}$ attains its maximum subject to $\| x\| = 1$ . Then find the unit vector $\vec {v}$ at which $Q$ attains its maximum subject to both $\| x\| =1$ and $\vec {x}\cdot \vec {u}=0$ .

$\vec {u} = \frac {1}{\answer {3}} \begin {bmatrix} \answer {2}\\ \answer {-1}\\ \answer {2} \end {bmatrix} \qquad \vec {v} = \frac {1}{\answer {3}} \begin {bmatrix} \answer {-1}\\ \answer {2}\\ \answer {2} \end {bmatrix}$ .