Let xn = Anx0. Then find x100 = A100x0. What happens to xn as n becomes very large?
Note: there is a typo in the exam, it is written xn = Anxn instead of xn = Anx0
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?
If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!
Don't calculate the 100th power of A directly. Instead, diagonalize A = TDT -1 and then take the 100th power of that.
Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work.
We are not going to multiply A with itself a hundred times, at least not directly. Instead, we are going to diagonalize A first, because the 100th power of a diagonal matrix is much easier to calculate. Indeed, if D is the diagonal matrix corresponding to A (i.e. is a matrix with the eigenvalues of A on the diagonal), and if T is the transformation matrix of eigenvectors of A, then
So let's find D, T and T -1. We need the eigenvalues and eigenvectors of A, so we calculate the roots of the characteristic polynomial :
And so the eigenvalues are 1/2, -1/3 and 1.
We compute an eigenvector for each eigenvalue using row reduction on the system .
For λ = 1 we have the system
And so this eigenspace is spanned by the eigenvector [2, 3, 3].
For λ = 1/2 we have the system
And so this eigenspace is spanned by the eigenvector [1, -1, 0].
For λ = -1/3 we have the system
And so this eigenspace is spanned by the eigenvector [2, 3, -5].
Using this basis of eigenvectors, we have that
We need now to compute the inverse of the matrix T. We obtain
So we can compute x100
As n becomes very large, xn approaches