Solving for complex eigenvalues

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Homework Help Overview

The discussion revolves around finding complex eigenvalues from a characteristic polynomial, specifically λ² + i = 0. Participants explore methods for solving this polynomial and the implications of complex roots in linear transformations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of deMoivre's theorem for solving the characteristic polynomial and express uncertainty about this method. Some inquire about alternative approaches, while others attempt to derive equations by equating real and imaginary parts of complex numbers.

Discussion Status

There is an ongoing exploration of different methods to solve for the eigenvalues, with some participants providing guidance on equating real and imaginary parts. Questions about the application of linear combinations in relation to the kernel of a transformation are also being examined, though clarity on these concepts remains elusive.

Contextual Notes

Participants note that they were not adequately shown methods for solving complex roots in class, which contributes to their uncertainty. There is also a mention of specific constraints regarding the transformation T and its kernel.

trap101
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Quick question:

I have a characteristic polynomial: λ2 + i = 0...how do I solve for the eigenvalues?

They're suppose to be + or - (√2/2)(1 - i) How'd they get those?
 
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trap101 said:
Quick question:

I have a characteristic polynomial: λ2 + i = 0...how do I solve for the eigenvalues?

They're suppose to be + or - (√2/2)(1 - i) How'd they get those?

λ^2=(-i)=exp(i*3*pi/2). Use deMoivre.
 
Last edited:
I'm kind of shaky with deMoivre, is there another way?
 
trap101 said:
I'm kind of shaky with deMoivre, is there another way?

Only harder ways. deMoivre is the easy way. One square root of exp(i*t) is exp(i*t/2). What does that lead to your case? Then use exp(i*t/2)=cos(t/2)+i*sin(t/2).
 
Only reason I ask is because we were not really shown the deMoivre way in class,...to think of it, we weren't even shown a way to solve for these complex roots. I'll give deMoivre's a try.

Another question, just out of curiosity:

Say I have a basis for the Ker(T). T being the transformation. Say I have an arbitrary vector v in V. Can I write v as a linear combination of the basis of the Ker(T), and if so do the co-efficients that I obtain from this linear combination make a co-ordinate vector in V, that if I apply the transformation T to this co-ordinate vector, will it map to the null-space in Im(T)?
 
trap101 said:
Only reason I ask is because we were not really shown the deMoivre way in class,...to think of it, we weren't even shown a way to solve for these complex roots. I'll give deMoivre's a try.

Another question, just out of curiosity:

Say I have a basis for the Ker(T). T being the transformation. Say I have an arbitrary vector v in V. Can I write v as a linear combination of the basis of the Ker(T), and if so do the co-efficients that I obtain from this linear combination make a co-ordinate vector in V, that if I apply the transformation T to this co-ordinate vector, will it map to the null-space in Im(T)?

Ok, for the first one you need (a+bi)(a+bi)=a^2-b^2+2abi=(-i). Equating real and imaginary parts you get two equations for the real numbers a and b. Now try to solve them.

For the second one. You can't write an arbitrary element of element of V as a combination of the basis of Ker(T) unless it's in Ker(T). And then it will map to 0 in Im(T). I'm not sure what the real question is here?
 
Dick said:
For the second one. You can't write an arbitrary element of element of V as a combination of the basis of Ker(T) unless it's in Ker(T). And then it will map to 0 in Im(T). I'm not sure what the real question is here?

Yup, that's what I was asking about.



Dick said:
Ok, for the first one you need (a+bi)(a+bi)=a^2-b^2+2abi=(-i). Equating real and imaginary parts you get two equations for the real numbers a and b. Now try to solve them.

so I get two equations: a2-b2 = 0 and 0 + 2abi = -i Is that what you mean?

I'm also trying the deMoivre way. So as of now, I'm converting -i into polar co-ordinates in order to use de Moivre: so currently I have 1 sin[itex]\sigma[/itex], how do you get [itex]\sigma[/itex] to be pi/2 ?
 
well I figured out the pi/2, how it's obtained; but I got -pi/2. I got it from b = r sin[itex]\sigma[/itex] ==>

-1 = 1 sin[itex]\sigma[/itex], and then figured out for [itex]\sigma[/itex].
 
wanted to delete this message...I'm actually still puzzled when it comes to the -pi/2
 
  • #10
trap101 said:
wanted to delete this message...I'm actually still puzzled when it comes to the -pi/2

exp(-i*pi/2)=cos(-pi/2)+i*sin(-pi/2)=(-i), right? A square root of that is exp(-i*pi/4)=cos(-pi/4)+i*sin(-pi/4). Which equals?
 

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