Find the eigenvalues and eigenvectors

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Homework Help Overview

The discussion revolves around finding the eigenvalues and eigenvectors of the matrix $$ A=\begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} $$, with a focus on the characteristic polynomial and the resulting system of equations for the eigenvectors.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to solve for eigenvectors after determining the eigenvalues using the characteristic polynomial. They express confusion regarding the resulting system of equations.
  • Some participants question the original poster's understanding of the dependency of the equations and suggest parametrizing the solutions.
  • Others discuss the validity of the eigenvector found and the verification process.

Discussion Status

The discussion has progressed with participants providing guidance on how to approach the system of equations. There is recognition of the linear dependence of the equations, and the original poster has acknowledged a misunderstanding in their calculations.

Contextual Notes

Participants are navigating through the implications of linear dependence in the context of eigenvectors and are addressing potential errors in the original poster's calculations.

Mutatis
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Homework Statement



Find the eigenvalues and eigenvectors fro the matrix: $$
A=\begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} $$.

Homework Equations



Characteristic polynomial: ## \nabla \left( t \right) = t^2 - tr\left( A \right)t + \left| A \right|## .

The Attempt at a Solution



I've found the eigenvalues doing through the characteristic polynomial equation above: $$ \lambda_1 = 1 $$ $$ \lambda_2 = -1 $$.
Then, to get the eigenvector associated to ## \lambda_1## the equation ##M v_1 = 0## must be satisfied, $$ \begin{pmatrix} \left(0-1\right) & -i \\ i & \left(0-1\right) \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = 0$$.
It leads me to a system that I'm having trouble to solve it: $$ \begin{cases} -x-iy=0 \\ ix-y=0 \end{cases} $$.
I don't know what to do next, please help me!
 
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Why are you having trouble ? What are you expecting to find ?
 
As I'm sure you realized, these equations are linearly dependent so the system has infinitely many solutions. So parametrize it, let ##y## be anything. Say ##y = a##. Solve for ##x##. That's a valid eigenvector. Any multiple of that is a valid eigenvector.
 
RPinPA said:
parametrize it, let yyy be anything. Say y=ay=ay = a. Solve for xxx. That's a valid eigenvector. Any multiple of that is a

I did what you've said ## y=1## then ##x=1/i=-i##, so I got ## v_1=(1, -i)##. When I put this vector in the matrix to verify ##Mv_1=0## it leads me to a non-zero value...
 
Mutatis said:
I did what you've said ## y=1## then ##x=1/i=-i##, so I got ## v_1=(1, -i)##. When I put this vector in the matrix to verify ##Mv_1=0## it leads me to a non-zero value...

That's odd: when I put ##x = -i## and ##y = 1## into ##-x - iy## I get ##0##, exactly as wanted. The second left-hand-side is just ##i \times## the first left-hand-side, so it will equal ##0## also.
 
Mutatis said:
I did what you've said ##y=1## then ##x=1/i=−i##, so I got ##v_1=(1,−i)##.

Are you sure? If ##x## is ##-i## and ##y## is ##1##, you would write that as ##(1, -i)##?
 
Oh! Guys I'm sorry I wrote the values wrong! Now I understand what I was doing wrong. Thank you very much guys!
 

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