# Complex eigenvalue proof

• Dusty912
What can you say about ##Y_1, Y_2## if this is true?In summary, we are trying to show that Y1 and Y2, with real entries, are linearly independent when A is a matrix with a complex eigenvalue λ=α+iβ. We start by expressing AY0=λY0 in terms of real numbers and vectors. Then, assuming Y1 and Y2 are linearly dependent, we get two equations by setting the real and imaginary parts of the resulting equation equal to each other. By plugging in the assumption that Y1=kY2, we can see that this would result in Y1 and Y2 being equal to 0, which contradicts the fact that they

## Homework Statement

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.

AY=λY

## 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 :)

Dusty912 said:

## Homework Statement

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.

AY=λY

## 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 :)
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##.

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

Dusty912 said:
okay so I would have A(Y1 + iY2)=(α+iβ)(Y1 + iY2)
and then I suppose I would multiply this out
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.)

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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?

Dusty912 said:
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?
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 (*).