# Thread: find an othonormal set of eigenvectors

1. ## find an othonormal set of eigenvectors

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.

2. Ok, where should you start? What have you done?

3. well i tried to find the characteristic equation,

and i got (x= lamda)
x^3 - 23x^2 - 168x - 396.. but then couldn't factorise it

4. Double-check your signs. I agree with the magnitudes of the coefficients, but I think you have a sign error in there somewhere.

5. ok, now i have , (x= lamda)

x^3 - 23x^2 + 168x - 396..

and i have (x-11)( x^2 -12x+36),

i think the second term is made up of two complex factors, but i forget how to do it?

6. Looking better. The quadratic factor you have there factors over the reals. In fact, it factors fairly straight-forwardly over the reals. You could always use the quadratic formula. What do you get?

7. lmao wow i can't believe i actually just did that. ( i swear i'm not this stupid, haha)
well, i have the eigeevalues now, i'l see how i go from here, thanks

8. Ok, let me know how it goes.

9. Originally Posted by linalg123
Find an orthonormal set of eigenvectors for the following symmetric matrix.

[ 10 0 2 ]
[ 0 6 0 ]
[ 2 0 7 ]
If you expand the determinant $\left|\begin{array}{ccc}10-\lambda & 0 & 2 \\ 0 & 6-\lambda & 0\\ 2 & 0 & 7-\lambda\end{array}\right|$ on the middle row you get $(6- \lambda)\left|\begin{array}{cc} 10- \lambda & 2 \\ 2 & 7- \lambda\end{array}\right|$ $= (6- \lambda)((10-\lambda)(7- \lambda)- 4)= (6- \lambda)(66- 17\lambda+ \lambda^2)$ which obviously has $\lambda= 6$ as a root.
Since the expansion you give obviously has 11 as a root, try factoring that quadratic as $(\lambda^2- 17\lambda+ 66)= (\lambda- 11)(\lambda- 6)$ 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.

10. ok, i have two linearly independant eigenvectors. Is that a problem that i can't get a third?

11. Show us your work. It isn't necessarily a problem. It depends on whether you're required to get a basis of orthonormal eigenvectors. If you're just asked to get a set of orthonormal eigenvectors, then who cares how many there are in the set?

12. 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?

13. that didn't work =S

14. It looks like you've correctly found two eigenvectors. (You should clean up the way that post looks by enclosing your LaTeX code with math tags.) However, you're missing an eigenvector.

15. 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?

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