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The VERY last part of finding an eigenvector

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)?

Thank you for your help?
Reply 1
Avatar for Hopple
Hopple
18
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.
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.
Reply 3
Avatar for Hopple
Hopple
18
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.
Reply 4
Avatar for TheEd
TheEd
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 [x−2x3]\begin{bmatrix} x \\ -\frac{2x}{3} \end{bmatrix} where x≠0x \neq 0

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