The VERY last part of finding an eigenvector

[claret_n_blue]

I can do everything, until I get to the point of actually putting the system of equations into that eigenvector "form". I won't use my actual numbers, but say I have solved everything and got:

ax = by
ax = by

Where a and b are 2 numbers, but a doesn't equal b.

Does this mean that for everyone 1 'y', I get (a/b) lots of x's, and so my vector is (a/b, 1). Or is it the other way round so for everyone 1 'x', I get (b/a) y's, so my vector is (1,b/a). Or is it simply (a,b)?

Thank you for your help :)

 

[AlephZero]

Oops, you got your algebra slightly wrong. It should be (b/a) lots of x's for every 1 y, etc.

The eigenvector just defines a "direction", not a magnitude, so you might as well use the simplest form you can write down. That would be (b,a) for your example.

For some applications of eigenvectors, the magnitude is "normalized", for example so that x^Tx = 1, or x^TAx = 1 where A is some matrix, or a particular element in the vector equals 1. But there is no point in doing that unless you have a reason for doing it.

 

[HallsofIvy]

I can do everything, until I get to the point of actually putting the system of equations into that eigenvector "form". I won't use my actual numbers, but say I have solved everything and got:

ax = by
ax = by

Where a and b are 2 numbers, but a doesn't equal b.

Does this mean that for everyone 1 'y', I get (a/b) lots of x's, and so my vector is (a/b, 1). Or is it the other way round so for everyone 1 'x', I get (b/a) y's, so my vector is (1,b/a). Or is it simply (a,b)?

Thank you for your help :)

How about "all of the above"? If you have a single equation, ax= by, then there are an infinite number of (x, y) pairs that satisfy the equation. (a/b, 1), (1, b/a), and (a, b) are all among them. Those all represent vectors in the same direction, or in the same subspace, with different lengths. So is another important pair: $(a/\sqrt{a^2+ b^2}, b/\sqrt{a^2+ b^2})$, the unit vector in that direction.

The crucial point is this- there is no single "eigenvector" corresponding to a given eigenvalue. The set of all eigenvectors corresponding to a given eigenvalue is a subspace. In particular, any multiple of an eigenvector is also an eigenvector. (a/b, 1)= (1/b)(a, b), (1, b/a)= 1/a(1, b), and $(a/\sqrt{a^2+ b^2}, b/\sqrt{a^2+ b^2})= [1/\sqrt{a^2+ b^2}](a, b)$ are all multiples of (a, b).