Is this normalised eigenvector undefined?

[ZedCar]

Homework Statement

I've started off with a 2x2 matrix of

(0) (i)
(-i) (0)

and I found the eigenvalues to be +1, -1

Then I found the resulting 2x1 eigenvectors to be

(-i)
(1)

and

(1)
(-i)

I now need the normalised eigenvectors.

Homework Equations

The Attempt at a Solution

I'm getting

{1/[sqrt(0)]} x eigenvector1
{1/[sqrt(0)]} x eigenvector2

Does this mean the normalised eigenvectors in this case are undefined?

 

[genericusrnme]

You need to take the dot product with the complex conjugate of the vector if it's over a complex field.

$|a| = \sqrt{a*\ .\ a}$

 

[ZedCar]

Thank you.

Do you mean then that the normalised eigenvectors would be

[1/(sqrt(a* . a))] x eigenvector1

and

[1/(sqrt(a* . a))] x eigenvector2

 

[genericusrnme]

yes, if a is the eigenvector in question.
This still holds if the eigenvectors are all real, the magnitude of a real vector a is still $\sqrt{a*\ .\ a}$ but since a is real a*=a and we get the familiar $\sqrt{a\ .\ a}$

 

[ZedCar]

I'm getting values for the normalised eigenvectors of

{1/[sqrt(2)]} x eigenvector1
{1/[sqrt(2)]} x eigenvector2

What do you think?

 

[genericusrnme]

That is correct!

 

[ZedCar]

Great!

I noticed there, with the two 2x1 eigenvectors which I in my first post being

(-i)
(1)

and

(1)
(-i)

For the number that's going to be under the radical, in this case we discovered it to be 2, can I simply take the [magnitude of (1)]^2 + the [magnitude of (-i)]^2 which equals 1+1=2 rather than take the dot product? Or am I effectively taking the dot product by doing this?

 

[HallsofIvy]

Yes, $(a+ bi)(a- bi)= a^2+ b^2$.

 

[genericusrnme]

You are taking the dot product when you do this, yes :p

 

[ZedCar]

That's great guys! Thanks very much for your help. It's much appreciated!

 

[ZedCar]

The final part of this question states that the original matrix is Hermitian, and that I should check that the eigenvalues and eigenvectors have the required properties for those of a Hermitian matrix.

The eigenvalues are both real, so that is satisfied.

Therefore for the eigenvectors to satisfy the required properties of a Hermitian matrix, (eigenvector1)^T x eigenvector2 = (eigenvector2)^T x eigenvector1 = 0

^T indicates Transpose

When I do this I'm getting -2i for both. So they are both equal to each other. Though 2i is not = 0 i.e. a null vector

 

[genericusrnme]

Remember that when you take the dot product of two complex vectors you need to take the complex conjugate of one of them.
You'll find that $e_1*.e_2 = e_2*.e_1 = 0$, where e1 and e2 are the two eigen vectors and * is complex conjugation.

 

[ZedCar]

That's great. Thanks very much again genericusrnme!

 

[genericusrnme]

No problem buddy