Solving Eigenvectors for Root -7 of 2x3 Matrix

[939]

Homework Statement

Find eigenvector for the root -7 of:

|2 3|
|3 -6|

Homework Equations

|2 3|
|3 -6|

The Attempt at a Solution

I got
1
-3

But my books says
-1
3

I am only wondering if this is possibly the same answer, because when I check my answer by multiplying the eigenvector by the original matrix and the root by the eigenvector the answer appears correct.

I.e.
|2 3|
|3 -6| times (1/-3) = (-7)(1/-3) = -7/21, while if you do the same for (-1/3), allbeit a different answer, but the condition still holds.

Is this correct? Is this condition truly the indicator of if you got the correct answer? (original vector * eigenvector) = (root * eigenvector)

 

[HallsofIvy]

Yes, any "eigenvector", v, corresponding to a given eigenvalue (I would not say "root") $\lambda$, of matrix A, has the property that $Av= \lambda v$.

What you are missing is that if $v$ is such an eigenvector then $av$, for any number a, is also an eigenvector, corresponding to eigenvalue $\lambda$: $A(av)= a (Av)= a(\lambda v)= \lambda (av)$.

In particular, with a= -1, if v is an eigenvalue then so is -v. The eigenvector is NOT unique.

 

[939]

Yes, any "eigenvector", v, corresponding to a given eigenvalue (I would not say "root") $\lambda$, of matrix A, has the property that $Av= \lambda v$.

What you are missing is that if $v$ is such an eigenvector then $av$, for any number a, is also an eigenvector, corresponding to eigenvalue $\lambda$: $A(av)= a (Av)= a(\lambda v)= \lambda (av)$.

In particular, with a= -1, if v is an eigenvalue then so is -v. The eigenvector is NOT unique.

Thanks a lot.

Just to confirm, the signs in this case do not matter and both answers are correct? I am such a noobie...