Question

Help with matrices:

If A= [ 1 -1 1]
........[ -1 0 0 ]
........[ 0 -1 1 ]
Find the trace of B= 2^(-98) * A^(100)

The problem here is to find A^(100), with elements in terms of 2.
Any idea, Thanks.

Answer 1

Try A·A and split the result in A+C, then
A·A·A=(A+C)·A=A+C+C·A (see what C·A looks like and find out that C·A=A+C too), then A·A·A=2(A+C). Now do:
A·A·A·A=2(A+C)·A=2A·A+2·C·A and we know this is: 2(A+C)+2·(A+C)=4(A+C)
Once again A·A·A·A·A=4(A+C)·A=4A·A+4·C·A=4(A+C)+4(A+C)=8(A+C)
In general,
\[A^n=2^{n-2}(A+C)\]
for n=100
\[B=2^{-98}·2^{98}·(A+C)=A+C\] then trace(B)=trace(A)+trace(C)

Answer 2

my idea is, try to diagonalize A by finding eigenvalue, eigenvectors and P , use that P to find D (diagonalized A) and then A^n = P-1 D^n P to get A^100, the leftover is easy, just time trace of A^100 by 2^-98
Hope this helps

Answer 3

@hoa
Totally agree with your approach. That was my first idea as it is the right method but found out that matrix A cannot be digonalized. Unless I have made a mistake in the calculations, it has two eigenvalues: λ=0 (order two) and λ=2 (single). When trying to get the base for eigenvectors I found out that for λ=0 the number of the base vectors was 1, that is lower than multiplicity (2 for λ=0), which is an indication, as you know, that the matrix cannot be diagonalized.

Answer 4

@CarlosGP. if you cannot diagonalize A to get D, how about J ? I mean Jordan Matrix we can take since we don't have D? We can apply the process on J exactly the way we apply to D. Hope it makes sense to you.

Answer 5

Might be done, and would be a more formal process than the pattern I have found. More tedious too, though