Why Are Eigenvectors with Complex Eigenvalues Linearly Independent?

[Dusty912]

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 :)

 

[Samy_A]

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 $Y_1, Y_2$.

 

[Dusty912]

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

 

[Samy_A]

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.)

 

[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 AY₁=αY₁-βY₂

but now what from here?

 

[Samy_A]

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 AY₁=αY₁-βY₂

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 (*).