# Finding eigenvector from eigenvalue

1. Mar 20, 2009

### DWill

1. The problem statement, all variables and given/known data
For the matrix A =

-1, 5
-2, -3

I found the eigenvalues to be -2 + 3i and -2 - 3i.
Now I need some help to find the eigenvectors corresponding to each.

2. Relevant equations

3. The attempt at a solution
For r = -2 + 3i, I plugged that into the (A - Ir) matrix, which I found to be

1-3i, 5
-2, -1-3i

I multiply that matrix with the vector (x y) and set it equal to (0 0) right? If I do that I get the following 2 equations:

(1-3i)x + 5y = 0
-2x - (-1-3i)y = 0

Did I make a mistake somewhere, or how should I go on to find the eigenvector? Thanks!

2. Mar 20, 2009

### Vuldoraq

Looks good to me so far, I haven't checked the actual numbers but the principles are fine.

Edit:Checked the numbers and they look fine too.

3. Mar 20, 2009

### HallsofIvy

Staff Emeritus
I prefer just to use the basic definition of "eigenvalue". If $\lambda$ is and eigenvalue of A then there exist a non-zero vector v such that $Av= \lambda v$ and, of course, v is an eigenvector.

Here, for $\lambda= -2+ 3}$ that is
$$A= \begin{bmatrix}-1 & 5 \\ -2 & -3\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}(-2+i\sqrt{3})x \\ (-2+3i)\end{bmatrix}$$
which reduces to two equations:
$-x+ 5y= (-2+ 3i)x$ and $-2x- 3y= (-2+ 3i)y$ both of which reduce to 5y= (-1+ 3i)x. Taking x= 5, y= -1+ 3i satisfies that and gives (5, -1+ 3i) as an eigenvector corresponding to eigenvalue -2+ 3i. A similar calculation gives an eigenvector corresponding to eigenvalue -2- 3i.