# Complex eigenvalue proof

1. Apr 16, 2016

### Dusty912

1. The problem statement, all variables and given/known data
Suppose the matrix A with real entries has the complex eigenvalue λ=α+iβ, β does not equal 0. Let Y0 be an eigenvector for λ and write Y0=Y1 +iY2 , where Y1 =(x1, y1) and Y2 =(x2, y2) have real entries. Show that Y1 and Y2 are linearly independent.
[Hint: Suppose they are not linearly independent. Then (x2, y2)=k(x1, y1[/SUB) for some constant k. Then Y0=(1+ik)Y1. Then use the fact that Y0 is an eigenvector of A and that AY1 contains no imaginary part.

2. Relevant equations
AY=λY

3. The attempt at a solution
Honestly, not too sure where to start for this one. I know I should begin by considering the scenario where Y1 and Y2 are not linearly independent, but I do not know where I should begin with this information. Thanks for your help :)

2. Apr 17, 2016

### Samy_A

You have $A Y_0 = \lambda Y_0$, $\lambda = \alpha +i \beta$ and $Y_0= Y_1+iY_2$.
Start by expressing $A Y_0 = \lambda Y_0$ in terms of the real numbers $\alpha, \beta$ and the real vectors $Y_1, Y_2$.

3. Apr 17, 2016

### Dusty912

okay so I would have A(Y1 + iY2)=(α+iβ)(Y1 + iY2)
and then I suppose I would multiply this out

4. Apr 17, 2016

### Samy_A

Yes, do that. Remember that $A, Y_1, Y_2, \alpha, \beta$ are all real. This will give you two equations (by setting the real and imaginary parts of the resulting equation equal to each other).
Then assume that $Y_1, Y_2$ are linearly dependent, and see what that gives. At this stage, remember that $\beta \neq 0$.

(Or you could reverse the order, first assume that $Y_1, Y_2$ are linearly dependent, and then do the multiplication.)

Last edited: Apr 17, 2016
5. Apr 17, 2016

### Dusty912

okay so the two equations I get (by setting the real and imaginary parts of the resulting equation equal to each other) is:
AiY2= iβY1 +αiY2 and AY1Y1Y2

but now what from here?

6. Apr 18, 2016

### Samy_A

So you have $AY_2=\beta Y_1 +\alpha Y_2$, $AY_1=\alpha Y_1 - \beta Y_2$ (*).

You want to prove that $Y_1, Y_2$ are linearly independent.
Assume that they are linearly dependent: that means that $\exists k \in \mathbb R, k\neq 0$ such that $Y_1=k Y_2$.
Plug this in into the equations (*).