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QUESTION

Write down the matrix $A$ corresponding to the quadratic form
$5x^2+5y^2+2z^2-2xy$. Find its eigenvalues and the corresponding
normalised eigenvectors, and write down a matrix $R$ such that
$R^TAR$ is diagonal with the eigenvalues of $A$ as its diagonal
entries. Show that this is so by calculating $R^TA$ and $(R^TA)R$.
Hence or otherwise diagonalise the quadratic form, i.e. write it
in the form $aX^2+bY^2+cZ^2$ giving explicit values for $a,b,c$
and give expressions for the new variables $X,Y,Z$ in terms of the
variables $x,y,z$.




ANSWER



$A=\left(\begin{array}{ccc}5&-1&0\\-1&5&0\\0&0&2\end{array}\right)$

$\det(A-\lambda I)=(2-\lambda)\left[(5-\lambda)^2-1\right]$ so
$\lambda=2$ or $(5-\lambda-1)(5-\lambda+1)=0\Rightarrow \lambda=2$
or $\lambda=4$ or $\lambda=6$.

Eigenvectors:

$\lambda=2$

$$\left(\begin{array}{ccc}3&-1&0\\-1&3&0\\0&0&0\end{array}\right)\mathbf{v}=\mathbf{0}$$

so $\left.\begin{array}{l}
3x=y\\x=3y\end{array}\right\}\Rightarrow x=y=0\ z=1\ \
\left(\begin{array}{c}0\\0\\1\end{array}\right)$


$\lambda=4$

$$\left(\begin{array}{ccc}1&-1&0\\-1&1&0\\0&0&-2\end{array}\right)\mathbf{v}=\mathbf{0}$$

 $\Rightarrow x=y\ z=0\ \
\left(\begin{array}{c}\frac{1}{\sqrt{2}}\\
\frac{1}{\sqrt{2}}\\1\end{array}\right)$


$\lambda=6$

$$\left(\begin{array}{ccc}-1&-1&0\\-1&-1&0\\0&0&-4\end{array}\right)\mathbf{v}=\mathbf{0}$$

 $\Rightarrow x=y=0\ z=1\ \
\left(\begin{array}{c}-\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}\\1\end{array}\right)$

Let
$R=\left(\begin{array}{ccc}\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}&0\\
\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}&0\\0&0&1\end{array}\right)$
so $R^T=R$

$$R^TA=\left(\begin{array}{ccc}\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}&0\\
\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}&0\\0&0&1\end{array}\right)
\left(\begin{array}{ccc}5&-1&0\\-1&5&0\\0&0&2\end{array}\right)=
\left(\begin{array}{ccc}2\sqrt{2}&2\sqrt{2}&0\\3\sqrt{2}&-3\sqrt{2}&0\\0&0&2
\end{array}\right)$$

$$(R^TA)R=\left(\begin{array}{ccc}2\sqrt{2}&2\sqrt{2}&0\\3\sqrt{2}&-3\sqrt{2}&0\\0&0&2
\end{array}\right)\left(\begin{array}{ccc}\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}&0\\
\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}&0\\0&0&1\end{array}\right)=\left(\begin{array}{ccc}
4&0&0\\0&6&0\\0&0&2\end{array}\right)$$

The quadratic form diagonalises to $4X^2+6Y^2+2Z^2$ where
$X=\frac{1}{\sqrt{2}}(x+y),\ Y=\frac{1}{\sqrt{2}}(x-y),\ z=Z$.



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