# Finding an eigenvector of a symmetric matrix given another eigenvector

• April 11th 2011, 08:45 AM
mbmstudent
Finding an eigenvector of a symmetric matrix given another eigenvector
Hi, I have the problem below and I don't know how to solve it... What property do symmetric matrices have that would allow me to solve it without any calculation? Thanks a lot!

So here is the problem:

Find an eigenvector for A (no calculation should be required: note that A is a symmetric matrix).

A =
[1, 1]
[1, -5]

and the eigenvector given is

[2]
[1]
• April 11th 2011, 09:25 AM
Deveno
(2,1) isn't an eigenvector. A(2,1) = (3,-3), which is not in span{(2,1)}. i am confused.
• April 11th 2011, 10:00 AM
FernandoRevilla
Quote:

Originally Posted by mbmstudent
So here is the problem: Find an eigenvector for A (no calculation should be required: note that A is a symmetric matrix).

A =
[1, 1]
[1, -5]

and the eigenvector given is

[2]
[1]

If the matrix is actually:

$A=\begin{bmatrix}{1}&{\;\;1}\\{1}&{-5}\end{bmatrix}$

then, its eigenvalues are $\lambda=-2\pm \sqrt{10}$ so, perhaps Ramanujan could find the eigenvectors with no calculation. :)
• April 11th 2011, 10:57 AM
zoek
Quote:

Originally Posted by mbmstudent
Hi, I have the problem below and I don't know how to solve it... What property do symmetric matrices have that would allow me to solve it without any calculation? Thanks a lot!

So here is the problem:

Find an eigenvector for A (no calculation should be required: note that A is a symmetric matrix).

A =
[1, 1]
[1, -5]

and the eigenvector given is

[2]
[1]

For a symmetric matrix it is known that:
eigenvectors corresponding to different eigenvalues are orthogonal.

If A has two different eigenvalues and [2, 1] is one of its eigenvectors then [-1, 2] is another one. (because their inner or dot product equals 0)
• April 11th 2011, 12:28 PM
Deveno
Quote:

Originally Posted by FernandoRevilla
If the matrix is actually:

$A=\begin{bmatrix}{1}&{\;\;1}\\{1}&{-5}\end{bmatrix}$

then, its eigenvalues are $\lambda=-2\pm \sqrt{10}$ so, perhaps Ramanujan could find the eigenvectors with no calculation. :)

• April 11th 2011, 12:41 PM
FernandoRevilla
Quote:

Originally Posted by Deveno