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Finding eigenvector from eigenvalue

  • Thread starter DWill
  • Start date
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1. Homework Statement
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. Homework 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!
 
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.
 

HallsofIvy

Science Advisor
Homework Helper
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I prefer just to use the basic definition of "eigenvalue". If [itex]\lambda[/itex] is and eigenvalue of A then there exist a non-zero vector v such that [itex]Av= \lambda v[/itex] and, of course, v is an eigenvector.

Here, for [itex]\lambda= -2+ 3}[/itex] that is
[tex]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}[/tex]
which reduces to two equations:
[itex]-x+ 5y= (-2+ 3i)x[/itex] and [itex]-2x- 3y= (-2+ 3i)y[/itex] 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.
 

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