Unitary Matrix, need to find eigen values/vectors

[orbitsnerd]

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!

 

[tiny-tim]

Hi orbitsnerd!

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! :yuck: have a lambda: λ and a square-root: √ )

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

λ - 1/√2  i/√2
 -1/√2    λ - i/√2

so what is the determinant of that? ​

 

[orbitsnerd]

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.

Hi orbitsnerd!


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

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

λ - 1/√2  i/√2
 -1/√2    λ - i/√2

so what is the determinant of that? ​

 

[tiny-tim]

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 …

just do it the usual way. ​