Find the eigenvalues and eigenvectors

  • #1
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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!
 

Answers and Replies

  • #2
dRic2
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Why are you having trouble ? What are you expecting to find ?
 
  • #3
RPinPA
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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.
 
  • #4
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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...
 
  • #5
Ray Vickson
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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.
 
  • #6
RPinPA
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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)##?
 
  • #7
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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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