Complex eigenvalue proof

  • Thread starter Dusty912
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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.

Homework Equations


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 :)
 

Answers and Replies

  • #2
Samy_A
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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.

Homework Equations


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##.
 
  • #3
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okay so I would have A(Y1 + iY2)=(α+iβ)(Y1 + iY2)
and then I suppose I would multiply this out
 
  • #4
Samy_A
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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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  • #5
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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?
 
  • #6
Samy_A
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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?
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 (*).
 

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