Are you saying that is a matrix such that there exists invertible such that
And you're asked to find
B is a 2x2 matrix with the property that there is an invertible 2x2 matrix R such that
R^-1BR = [(sqrt 3) 0 ] [ 1.2 -(sqrt 3)/2]
[ 0 (sqrt 3) ] [(sqrt 3)/2 1/2 ]
I know the first part is the scaling matrix and the second on in the rotation matrix.
I'm suppose to find B^6 which is equal to
B^6 = R^-1D^6R
But I dont know how I would find the matrix for R or D.
Hmm. Interesting problem. Well, the nice thing about diagonal matrices is that it's easy to find their nth power: just take the nth power of the elements on the main diagonal. Now the problem you've been given is not a typical diagonalization problem. Presumably, you've seen those already, so you know how you can always do this:
If
then
I played some "what if" games with your situation. Let's say you've got
So, is known. Then
and the is on the wrong side of We want to have
If I compare these two, it leads me to ask this question: what if I could find a diagonal such that
Then
as desired.
So, can you diagonalize but then, when you find the invertible matrix can you take its square root? There are ways to do that, I think, that aren't overly difficult, especially if you orthogonally diagonalize
So, give this program a shot: orthogonally diagonalize What do you get?
Are you sure about that? I recommend you go back and reread the problem. I am inclined to suspect that it is R itself that is equal to and that then
which is then the same as saying that .
If that is the case, that you have , where I am using "D" to stand for that diagonal matrix, then you also have . Now notice that
, etc.
Further, since D is a diagonal matrix, it is easy to find powers of D:
.