bb7c8f…: “Less a proud dodecahedron”

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Compute the quadratic form $Q(\vec {x})= \vec {x}^TA\vec {x}$ given $\vec {x} =\begin {bmatrix} x_1\\ x_2 \end {bmatrix}$ and $A = \begin {bmatrix} 2&6\\6&1\end {bmatrix}$ .

$Q(\vec {x}) = \answer {2}x_1^2 +\answer {1}x_2^2 +\answer {12}x_1x_2$

Compute the matrix of the quadratic form:
$Q(\vec {x}) = 7x_1^2 +3x_2^2 +8x_1x_2$ . Assume $\vec {x} \in \RR ^2$ .

$A = \begin {bmatrix} \answer {7} & \answer {4}\\ \answer {4} & \answer {3}\end {bmatrix}$

Compute the matrix of the quadratic form:
$Q(\vec {x}) = x_1^2 -x_2^2 +6x_3^2 +4x_1x_2-10x_1x_3$ . Assume $\vec {x} \in \RR ^3$ .

$A = \begin {bmatrix} \answer {1} & \answer {2}& \answer {-5}\\ \answer {2} & \answer {-1}& \answer {0}\\ \answer {-5}& \answer {0}& \answer {6}\end {bmatrix}$

Suppose $P$ orthogonally diagonalizes $A$ . Then $A=PDP^{-1}$ for some diagonal matrix $D$ . Which of the following is equivalent to $P^TAP$ ?

$D$ $PDP^{-1}$ $P^TDP$ cannot be simplified

If $P$ orthogonally diagonalizes $A$ , then $P$ is an orthogonal matrix which means $P^T=P^{-1}$ .

Suppose $P$ orthogonally diagonalizes $A$ . Substitute $\vec {x} = P\vec {y}$ into the Quadratic form $Q(\vec {x}) = \vec {x}^TA\vec {x}$ and simplify as much as possible. This substitution is called a change of variable.

$P\vec {y}^TAP\vec {y}$ $\vec {y}^TD\vec {y}$ $P^T\vec {y}^TAP\vec {y}$ $A\vec {y}$ $D\vec {y}$

The previous question might be helpful.

The orthogonal diagonalization of $A$ is given below. Substitute the change of variable $\vec {x} = P\vec {y}$ into the Quadratic form $Q(\vec {x}) = \vec {x}^TA\vec {x}$ and give the new quadratic form. $A = PDP^{-1} \text { where } P = \begin {bmatrix} -2/3 &-1/3&2/3\\ -2/3&2/3&-1/3\\ 1/3& 2/3& 2/3\end {bmatrix} \text { and } D= \begin {bmatrix} 3&0&0\\0&9&0\\0&0&15\end {bmatrix}$

The previous question might be helpful.

The new quadratic form is $\answer {3}y_1^2 +\answer {9}y_2^2+\answer {15}y_3^2$ . Notice that it has no cross product terms. i.e. no $y_iy_j$ where $i\neq j$ .