- Problem Statement
- Find eigenvalues & eigenvectors
- Relevant Equations
- det(A-r*I) = 0
Thanks for your time.
Ahh, I see. Thank youSort of. Those are the same two eigenvectors up to a constant, but in opposite order.
Multiply your ##\vec v## by the constant (1 + i). As you know, any multiple of an eigenvector is still an eigenvector of the same eigenvalue.
$$(1 + i) {2 \choose {2 - 2i}} = {{2 + 2i} \choose {2(1 - i)(1+i)}} = {{2+2i} \choose 4}$$
and similarly multiply ##\vec w## by the constant (1 - i).
$$(1 - i) {2 \choose {2 + 2i}} = {{2 - 2i} \choose {2(1 + i)(1-i)}} = {{2-2i} \choose 4}$$
Since the order is reversed, so are the eigenvalues with which each is associated.