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QUESTION

For each of the following matrices $A$ find an orthogonal matrix
$P$ such that $P^tAP$ is diagonal:

$$ \left[\begin{array}{rr}
 10 & 2 \\
  2 & 7 \\
\end{array}\right]
\makebox[1cm]{} \left[\begin{array}{rrr}
 4 & 2 & 1 \\
 2 & 3 & 2 \\
 1 & 2 & 4 \\
\end{array}\right]
\makebox[1cm]{} \left[\begin{array}{rrrr}
 0 & 1 & 1 & 1 \\
 1 & 0 & 1 & 1 \\
 1 & 1 & 0 & 1 \\
 1 & 1 & 1 & 0 \\
\end{array}\right].
$$

\bigskip

ANSWER



\begin{tabular}{lclc}
First matrix:\\
Eigenvalue&6&normalised
eigenvector&$\left[\begin{array}{c}\frac{1}{\sqrt{5}}\\-\frac{2}{\sqrt{5}}\end{array}\right]$\\
Eigenvalue&11&normalised
eigenvector&$\left[\begin{array}{c}\frac{2}{\sqrt{5}}\\\frac{1}{\sqrt{5}}\end{array}\right]$\\
\\
Second matrix:\\
Eigenvalue&1&normalised
eigenvector&$\left[\begin{array}{c}\frac{1}{\sqrt{6}}\\-\frac{2}{\sqrt{6}}\\\frac{1}{\sqrt{6}}\end{array}\right]$\\
Eigenvalue&3&normalised
eigenvector&$\left[\begin{array}{c}\frac{1}{\sqrt{2}}\\0\\-\frac{1}{\sqrt{2}}\end{array}\right]$\\
Eigenvalue&7&normalised
eigenvector&$\left[\begin{array}{c}\frac{1}{\sqrt{3}}\\\frac{1}{\sqrt{3}}\\\frac{1}{\sqrt{3}}\end{array}\right]$\\
\end{tabular}

\medskip
Third matrix:

 The slick way to do the question is as follows.
\medskip

Clearly $\lambda=-1$ is an eigenvalue but rank $(A+I)=1$, so
null$(A+I)=3$. Hence $-1$ is a triple eigenvalue and (as the
matrix is symmetric) the eigenspace is three-dimensional. The
other eigenvalue is 3 (use the fact that the sum of the
eigenvalues equals the trace or notice that the sum of entries in
each row equals 3).

The eigenvalue $-1$ has eigenspace $w+x+y+z=0$ so three mutually
orthogonal eigenvectors in this space are required. In general one
can choose any three independant eigenvectors and then use
Gram-Schmidt to get an orthogonal set, but in the present case it
is not hard to spot three orthogonal vectors.

Eigenvalue $-1$; normalised eigenvectors
$\left[\begin{array}{c}\frac{1}{2}\\\frac{1}{2}\\-\frac{1}{2}\\-\frac{1}{2}\end{array}\right],
\left[\begin{array}{c}\frac{1}{2}\\-\frac{1}{2}\\-\frac{1}{2}\\\frac{1}{2}\end{array}\right],
\left[\begin{array}{c}-\frac{1}{2}\\\frac{1}{2}\\-\frac{1}{2}\\\frac{1}{2}\end{array}\right].$

The eigenvalue 3 has normalised eigenvector
$\left[\begin{array}{c}\frac{1}{2}\\\frac{1}{2}\\\frac{1}{2}\\\frac{1}{2}\end{array}\right]$



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