Question

I am working with complex eigenvalues and I noticed that depending on the equation I choose from A-lambda(I) = 0 I could get two different eigenvectors. I would like to know why and if there is a correct eigenvector and I can pick either one. For clarification, my class did not cover unit eigenvectors and we are not expected to produce unit eigenvectors on the final.

3.4 p) 14 l. Corpute egen values dt 2. -小 2 2 472 -3 lo Can I hase 2 dPprent eiscnvectous?

Answer 1

Please check the step, here you have made mistake 4 3+47i 2. 4- 2. 4- You made mistake in the element (1,1) of the matrix, 1 - V47I [ 1 t2Hİ) and not-AM) herepore, 2 1-,47 I 2

so, if yo-6, then r,-1-1471 From eq(1), -(1+147 j ) χ。zfy。 y,-+ (1+147,) Xo=8, then SO, İF Thus ev,- Multipy ev, by 1+ 4i ie (1+47 i) 6 1- (-47) 6 Tris implies, So, you can pick either of ev or e since they are stant multiple of one another ie. ev is linearly dependern

on ev2. That's why you can pick either one as one aiways be a constant multiple of akways be a canstant multiple o another.