Unitary Matrix, need to find eigen values/vectors

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Homework Help Overview

The discussion revolves around finding the eigenvalues and eigenvectors of a given unitary matrix that contains complex numbers. The original poster expresses difficulty in solving for these eigenvalues and vectors due to the presence of imaginary numbers.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the determinant of a modified matrix derived from the original matrix, questioning how to simplify the calculation. There are attempts to identify factors of the determinant and to isolate relevant entries for easier computation.

Discussion Status

Some participants have provided insights into the determinant and its factors, which may guide the original poster in their calculations. However, there is still uncertainty regarding the process of finding eigenvectors, particularly with the involvement of imaginary numbers.

Contextual Notes

The original poster mentions needing to plug eigenvalue results back into the original matrix to find eigenvectors, indicating a standard approach is being followed. The presence of imaginary numbers is noted as a source of confusion.

orbitsnerd
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Homework Statement



matrix:

1/sqrt(2) i/sqrt(2) 0

-1/sqrt(2) i/sqrt(2) 0

0 0 1

Find eigen values and eigen vectors and determine if it is diagonalizable



Homework Equations



The matrix is unitary because Abar*Atranspose=I (identity matrix)




The Attempt at a Solution



I am having problems solving for the eigenvalues and vectors because of the imaginary numbers. What I get is:

lambda-1/sqrt(2) i/sqrt(2) 0

-1/sqrt(2) lambda- i/sqrt(2) 0

0 0 lambda-1


=(lambda-1/sqrt(2))*(lambda- i/sqrt(2))*(lambda-1)-(i/sqrt(2))*(-1/sqrt(2))*(lambda-1)

I need help getting to the next step.
Thanks!
 
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Hi orbitsnerd! :smile:
orbitsnerd said:
lambda-1/sqrt(2) i/sqrt(2) 0

-1/sqrt(2) lambda- i/sqrt(2) 0

0 0 lambda-1


=(lambda-1/sqrt(2))*(lambda- i/sqrt(2))*(lambda-1)-(i/sqrt(2))*(-1/sqrt(2))*(lambda-1)

(euuugh! have a lambda: λ and a square-root: √ :wink:)

Look at it … (λ - 1) is obviously a factor of the determinant, so you can ignore everything except the four top-left entries:

Code:
λ - 1/√2  i/√2
 -1/√2    λ - i/√2

so what is the determinant of that? :smile:
 
Awesome short cut. I now have my eigenvalues as:

λ1=1, λ2=(1+√3)/(2√2) + [(1-√3)/(2√2)]i and λ3=(1-√3)/(2√2) + [(1+√3)/(2√2)]i

I have issues finding the eigenvectors. I know you need to plug in the values of each λ back into the original matrix and solve for e1, e2 and e3. The imaginary number throws me off in this case.


tiny-tim said:
Hi orbitsnerd! :smile:


(euuugh! have a lambda: λ and a square-root: √ :wink:)

Look at it … (λ - 1) is obviously a factor of the determinant, so you can ignore everything except the four top-left entries:

Code:
λ - 1/√2  i/√2
 -1/√2    λ - i/√2

so what is the determinant of that? :smile:
 
orbitsnerd said:
I have issues finding the eigenvectors. I know you need to plug in the values of each λ back into the original matrix and solve for e1, e2 and e3. The imaginary number throws me off in this case.

I don't see what the problem is :confused:

just do it the usual way. :smile:
 

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