So, im doing some vector algebra and im doing some studing on eigenvalues and eigenvectors

1)general calculation of eigenvalues

the question is
find the eigenvectors for the 2x2 matrix
(3,2)
(2,3)

anyway, i found the eigenvalues were 5 and 1. i found the eigenvector for 5 is (1,1) which is right, but for the eigenvalue 1...
i find the the value of the matrix A,minus the product of the eigenvalue (1) and the identity matrix, then find the solution for it when it equals the zero vector on an augmented matrix
(2,2|0)
(2,2|0) row reducing, i get

(2,2|0)
(0,0|0)
let the second column be y,
2*x + 2*y = 0
x=-1
therefore, the eigenvector must be (-1,1). but the answer tells me its (1,-1)

whats up with that


2) complex eigenvalues

ok. so the question is
find the eigenvectors for the 2x2 matrix
A =
(1,2)
(-2,1)

i found the eigenvalues were complex
=1+2i, 1-2i
to find the eigenvalue for 1+2i, i get the matrix A, and subtract it by the product of the eigenvector and the identity matrix, put it in an augmented matrix and equate them to the zero vector. this gives
(-2i,2|0)
(-2,-2i|0)
row reducing gives me
(-2i,2|0)
(0,0|0)
now i let the second column variable be y. therefore, by back substitution,
-2i*x + 2*y = 0
solving for x,
x = y/i

so shouldnt the eigenvector for 1+2i be (-1/i, 1)? the answer im given is (-i/1, 1)
so what do i do?

Any constant multiple of an eigenvector of a particular eigenvalue is also an eigenvector. So for instance, in 1), since you found your eigenvector to be (-1,1), then -1*(-1,1)=(1,-1) is also an eigenvector.

If you need to justify this fact in your homework or whatever, its easy to do using the definition of an eigenvector and the property of any matrix A that A(cx)=cAx for a constant c. Say x is an eigenvector of A with eigenvalue k. Then Ax=kx (by definition of an eigenvector). Let c be any constant. Then A(cx)=cAx=ckx=k(cx). So since A(cx)=k(cx), then cx is an eigenvector of A with eigenvalue k, just as x was.

(1) Multiplied by -1, both valid answers. I usually normalize to stay consistent.

(2) Divide the first row through by 2 once you do your row reduction. This equation tells you that any vector that solves (-i)x1 + (1)x2 = 0 is an eigenvector, so the components can just be -i and 1. Conjugate is the second eigenvector.