You would solve for the eigenvalues by solving the numeric equation . If we let , then we have which, by the quadratic formula, has roots . The eigenvalues of the original equation are the roots of those and so will be very messy!
Let M = { [a -b] , a,b E R }
..............[b a]
Solve X^4+X^2+I=0, where X E M
I attempted this question a while ago. I remember finding the eigenvalues and eigenvectors (X^4+X^2=-I and treat X^4+X^2 as a matrix itself) but it led to nowhere. I then tried to use a linear map but no luck either.
I would love to hear ideas on how to solve this one.
Thanks.
Where and .
So M is a rotation by with a scaling by .
Immediately using a geometric interpretation, we can see that a rotation by 60 degrees is a solution for X. Also a rotation by 120 degrees is another solution.
There are more solutions, but hopefully this helps you get started finding roots.
An alternative: we can use the natural field isomorphism:
so, solving
in
it is equivalent to solve
in .
But,
As a consequence the roots of the given equation are the elements of the set:
Fernando Revilla
Thanks for your help guys.
FernandoRevilla, I haven't learn anything about natural field isomorphism so I don't understand what you did there...Is that some kind of a linear map? Is there any link about it so I can have a read? I tried google but nothing really came up.
EDIT: you can actually prove that its a linear map. =) Thanks that's a very pretty solution!
snowtea, could you please explain how you got 60 as a solution using geometric interpretation? I didn't quite see that =S
I get everything up to "So M is a rotation by with a scaling by gamma." X^4 and X^2 mean rotations by 4*theta and 2*theta respectively(with scalings too). But what about X^4+X^2+I? How did you see that 60 is a solution so quickly?
Let X be a rotation by 60 degrees (no scaling).
Draw a vector v.
Multiplying by I keeps it the same.
Multiplying it by X^2 rotates it by 120 degrees.
Multiplying it by X^4 rotates it by 240 degrees.
(X^4+X^2+I) v is the sum of all the above vectors.
You can show this is always 0 using some simple trigonometry.