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 #3
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Mmmm....let's look at this one again....

We need to solve 

(A− I)u=0
 
I'm going to let x = u 1   and y = u 2.......Also, I don't know how to represent vectors in this thing, but I think you can follow what  I'm doing
 
Going from this point forward, we get

1 -2i    -5                      x       =          0

1        -1 - 2i                 y       =          0

This generates the following system

(1 - 2i) x - (5)y          = 0

1x           -(1 + 2i)y    = 0

Using the second equation, we have

x = (1 + 2i)y

And if we let y = some convenient value, say, 1....then x  = (1 + 2i)  and we have the following eigenvector associated with   lambda = (1 + 2i)

(1 + 2i)

     1

BTW....you can check this solution in the 1st equation of the system.....I believe it "works"

 

To find the eigenvector associated with the second lambda, (1 - 2i), we follow a similar process and we have

1 +2i    -5                      x       =          0

1        -1 + 2i                 y       =          0

This generates the following system

(1 + 2i) x - (5)y       = 0

1x           -1y + 2iy  = 0

And using the second equation, we have

x = (1 - 2i)y

And if we let y = some convenient value, say, 1....then x  = (1 - 2i)  and we have the following eigenvector associated with   lambda = (1 - 2i)

(1 - 2i)

     1

As before....you can check this solution in the 1st equation of the system.....

I think this is correct, but my Linear Algebra skills have been sitting in the closet for awhile...

17.04.2014