Unitary Matrix, need to find eigen values/vectors

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:
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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