helloo!
find the eigenvalues and eigen vectors of
A= |1 2|
|5 4|
i dont think its long to work out, but ive just started this topic and cannot work it out
A non zero vector x is called an eigenvector of A if Ax is a scalar multiple of x.
Where k is the eigenvalue.
They're tied together. Lambda is mostly used, but I will use k.
You know your eigenvalues, -1 and 6. They're the roots of your characteristic polynomial.
In the first matrix, we have
Let y=t, then we have x=-t and y=t
In the second one, we have
Let y=t, then
Therefore, the eigenvectors are
and
i understand how to get upto here, but i dont see how you would get the eigenvectors because the teacher didnt explain the steps. i thought that for the 2255 matrix, it would be 2X1+2X2=0 so x1=-X2 so in the eigenvector, the top coordinate would be -1 and the same principle works for the bottom so it would be -1 again so the eigenvector would be (-1/-1) but i think this is wrong...
Let's try this in a different way. We know the eigenvalues are -1 and 6. So we may write the equation for the -1 eigenvector as (from the eigenvalue equation ):
So we have the simultaneous equations
and
Solving this for a and b gives a = a and b = -a. So the eigenvector is
(This is just the negative of galactus' form for the eigenvector. Since the variable a is unspecified we may easily set t = -a and get my form from his.)
We may get an unambiguous eigenvector if we have some condition on the eigenvector, such as orthonormality or something. Since this wasn't mentioned I will leave the eigenvector like this. The 6 eigenvalue is similar:
I'll leave it to you to work out the rest.
-Dan