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]

(2,1) isn't an eigenvector. A(2,1) = (3,-3), which is not in span{(2,1)}. i am confused.

Quote:

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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:




then, its eigenvalues are so, perhaps Ramanujan could find the eigenvectors with no calculation. :)

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)

Quote:

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Originally Posted by FernandoRevilla

If the matrix is actually:




then, its eigenvalues are so, perhaps Ramanujan could find the eigenvectors with no calculation. :)

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no he couldn't...he's dead.

Quote:

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Originally Posted by Deveno

no he couldn't...he's dead.

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Thanks for the information.

do i detect a note of sarcasm?

in any case, the original problem as posed is ill-formed. i urge the original poster to check his input.