Finding A to the power of n without using eigenvalues

[dlevanchuk]

Oh it gives me headache... been thinking on this problem for a while, and don't even know where to begin! Could anyone give me a hint at least?? :(

Problem:
Let A be (3x3) matrix : [ 4 -2 2; 2 4 -4; 1 1 0] and u (vector) = [1 3 2].
a) Verify that Au = 2u
I got this one without a problem.

b) Without forming A^5, calculate the vector A^5*u.
This is where I get stuck.. I've tried to search, but keep coming up with some equations that involve eigenvalues (which I haven't studied yet..). So, I am assuming that i don't have to use any of eigenvalues.. Is there any other way?
I tried to replace A matrix with [a b c; d...] values, and take first 3 powers of that, but its way too hard to keep track of everything..

So, any hint?? :(

 

[CompuChip]

You know that Au = 2u.

Now what is A²u = A (A u)?

Note that numbers commute with matrices, so for example A(2u) = 2(Au).

Actually you are using eigenvalues here: when Av = c v for some (non-trivial) vector v and some number c, then we say that v is an eigenvector for the matrix A, with corresponding eigenvalue c.[/size]

 

[dlevanchuk]

Oh man! Thats simple! Didn't even think about relating problem a with problem b! Thanks a ton!