Find the eigenvalues and eigenvectors

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The discussion focuses on finding the eigenvalues and eigenvectors of the matrix A = [[0, -i], [i, 0]]. The eigenvalues calculated are λ1 = 1 and λ2 = -1. The user struggles with solving the system of equations derived from the eigenvalue equation, which is linearly dependent, indicating infinitely many solutions. After guidance, the user successfully parametrizes the equations and finds an eigenvector v1 = (1, -i). The user realizes a mistake in their previous calculations and expresses gratitude for the assistance received.
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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