# Solving for complex eigenvalues

Quick question:

I have a characteristic polynomial: λ2 + i = 0.........how do I solve for the eigenvalues?

They're suppose to be + or - (√2/2)(1 - i) How'd they get those?

Dick
Homework Helper
Quick question:

I have a characteristic polynomial: λ2 + i = 0.........how do I solve for the eigenvalues?

They're suppose to be + or - (√2/2)(1 - i) How'd they get those?

λ^2=(-i)=exp(i*3*pi/2). Use deMoivre.

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I'm kind of shaky with deMoivre, is there another way?

Dick
Homework Helper
I'm kind of shaky with deMoivre, is there another way?

Only harder ways. deMoivre is the easy way. One square root of exp(i*t) is exp(i*t/2). What does that lead to your case? Then use exp(i*t/2)=cos(t/2)+i*sin(t/2).

Only reason I ask is because we were not really shown the deMoivre way in class,...to think of it, we weren't even shown a way to solve for these complex roots. I'll give deMoivre's a try.

Another question, just out of curiosity:

Say I have a basis for the Ker(T). T being the transformation. Say I have an arbitrary vector v in V. Can I write v as a linear combination of the basis of the Ker(T), and if so do the co-efficients that I obtain from this linear combination make a co-ordinate vector in V, that if I apply the transformation T to this co-ordinate vector, will it map to the null-space in Im(T)?

Dick
Homework Helper
Only reason I ask is because we were not really shown the deMoivre way in class,...to think of it, we weren't even shown a way to solve for these complex roots. I'll give deMoivre's a try.

Another question, just out of curiosity:

Say I have a basis for the Ker(T). T being the transformation. Say I have an arbitrary vector v in V. Can I write v as a linear combination of the basis of the Ker(T), and if so do the co-efficients that I obtain from this linear combination make a co-ordinate vector in V, that if I apply the transformation T to this co-ordinate vector, will it map to the null-space in Im(T)?

Ok, for the first one you need (a+bi)(a+bi)=a^2-b^2+2abi=(-i). Equating real and imaginary parts you get two equations for the real numbers a and b. Now try to solve them.

For the second one. You can't write an arbitrary element of element of V as a combination of the basis of Ker(T) unless it's in Ker(T). And then it will map to 0 in Im(T). I'm not sure what the real question is here?

For the second one. You can't write an arbitrary element of element of V as a combination of the basis of Ker(T) unless it's in Ker(T). And then it will map to 0 in Im(T). I'm not sure what the real question is here?

Ok, for the first one you need (a+bi)(a+bi)=a^2-b^2+2abi=(-i). Equating real and imaginary parts you get two equations for the real numbers a and b. Now try to solve them.

so I get two equations: a2-b2 = 0 and 0 + 2abi = -i Is that what you mean?

I'm also trying the deMoivre way. So as of now, I'm converting -i into polar co-ordinates in order to use de Moivre: so currently I have 1 sin$\sigma$, how do you get $\sigma$ to be pi/2 ?

well I figured out the pi/2, how it's obtained; but I got -pi/2. I got it from b = r sin$\sigma$ ==>

-1 = 1 sin$\sigma$, and then figured out for $\sigma$.

wanted to delete this message.......I'm actually still puzzled when it comes to the -pi/2

Dick