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# The VERY last part of finding an eigenvector Watch

1. I can do everything, untill 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

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)?

2. It might be clearer if you did use your numbers. x is an eigenvector if Ax=cx, where A is a matrix, x a vector and c a scalar.

Is your 'by' a scalar multiple of x? If so, and if a is your matrix, then you x is an eigenvector.
3. (Original post by Hopple)
It might be clearer if you did use your numbers. x is an eigenvector if Ax=cx, where A is a matrix, x a vector and c a scalar.

Is your 'by' a scalar multiple of x? If so, and if a is your matrix, then you x is an eigenvector.
If I use numbers, this is what has happened:

I have the matrix:

10 -6
-6 5

I worked out the eigenvalues to be 14 and 1 and then to work out the eigenvector when lambda = 14, I get

(A - lambda I). v = 0
v = (x,y)

I get -4x - 6y = 0
and -6x - 9y = 0

So by rearranging and simplifying, I get

-2x = 3y
-2x = 3y

Now I need to put this into the vector form thing, but I don't get what you have to do.
4. That shows your equations are consistent. Pick a number for x - let's pick 3 (I picked 3 to make the numbers nice, but any number will do), so y=-2. An eigenvector is then (3,-2). Picking different values for x will give you a scalar multiple of that eigenvector.
5. (Original post by claret_n_blue)
If I use numbers, this is what has happened:

I have the matrix:

10 -6
-6 5

I worked out the eigenvalues to be 14 and 1 and then to work out the eigenvector when lambda = 14, I get

(A - lambda I). v = 0
v = (x,y)

I get -4x - 6y = 0
and -6x - 9y = 0

So by rearranging and simplifying, I get

-2x = 3y
-2x = 3y

Now I need to put this into the vector form thing, but I don't get what you have to do.
This shows that any eigenvector of A with eigenvalue 14 has the form where

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Updated: December 8, 2010
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