Find an orthonormal set of eigenvectors for the following symmetric matrix.
[ 10 0 2 ]
[ 0 6 0 ]
[ 2 0 7 ]
that's a 3x3 matrix, just don't know how to put it all in one.
If you expand the determinant $\displaystyle \left|\begin{array}{ccc}10-\lambda & 0 & 2 \\ 0 & 6-\lambda & 0\\ 2 & 0 & 7-\lambda\end{array}\right|$ on the middle row you get $\displaystyle (6- \lambda)\left|\begin{array}{cc} 10- \lambda & 2 \\ 2 & 7- \lambda\end{array}\right|$$\displaystyle = (6- \lambda)((10-\lambda)(7- \lambda)- 4)= (6- \lambda)(66- 17\lambda+ \lambda^2)$ which obviously has $\displaystyle \lambda= 6$ as a root.
Since the expansion you give obviously has 11 as a root, try factoring that quadratic as $\displaystyle (\lambda^2- 17\lambda+ 66)= (\lambda- 11)(\lambda- 6)$ so that 6 is a double eigenvalue and 11 is an eigenvalue. Now start looking for eigenvectors.
(Double click on the formulas to see the LaTex code.)
that's a 3x3 matrix, just don't know how to put it all in one.
for \lamda =11
\left|\begin{array}{ccc}1 & 0 & -2 \\ 0 & 5 & 0\\ 2 & 0 & 4\end{array}\right|
y=0
x=2z eigenvector \left|\begin{array}{ccc}2 & 0 & 1\end{array}\right|
for \lamda=6
\left|\begin{array}{ccc}-4 & 0 & -2 \\ 0 & 0 & 0\\ -2 & 0 & -1\end{array}\right|
y=0
x=-2z eigenvector \left|\begin{array}{ccc}1 & 0 & -2\end{array}\right|
that look ok? so then i just make them orthonormal and i'm done?
for $\displaystyle \lamda =11$
$\displaystyle
\left|\begin{array}{ccc}1 & 0 & -2 \\ 0 & 5 & 0\\ 2 & 0 & 4\end{array}\right|$
y=0
x=2z eigenvector $\displaystyle \left|\begin{array}{ccc}2 & 0 & 1\end{array}\right|$
for $\displaystyle \lamda=6$
$\displaystyle \left|\begin{array}{ccc}-4 & 0 & -2 \\ 0 & 0 & 0\\ -2 & 0 & -1\end{array}\right|$
y=0
x=-2z eigenvector $\displaystyle \left|\begin{array}{ccc}1 & 0 & -2\end{array}\right|$
that look ok? so then i just make them orthonormal and i'm done?