No.
In order to solve the problem in an efficient way, I think the use of eigenvalues and eigenvectors are essential. The concept and use of these quantities are in themselves a subject of study. I think
https://www.math.hmc.edu/math40-01/math40-lect14.pdf
makes for a good introduction to eigenvalue and eigenvectors.
I'll do a brief discussion and use 2X2's for examples to keep things simple: Let
A = \(\displaystyle \begin{pmatrix}
a_{11}& a_{12}\\ a_{21}& a_{22}\end{pmatrix}\)
The idea behind raising a matrix to a large power is that, for example, we can write
A
2 = \(\displaystyle A \begin{pmatrix}
a_{11}& 0\\ a_{21}& 0\end{pmatrix}\, +\, A \begin{pmatrix}
0& a_{12}\\ 0& a_{22}\end{pmatrix}\) = \(\displaystyle \begin{pmatrix}
b_{11}& b_{12}\\ b_{21}& b_{22}\end{pmatrix}\)
A
3 = \(\displaystyle A \begin{pmatrix}
b_{11}& 0\\ b_{21}& 0\end{pmatrix}\, +\, A \begin{pmatrix}
0& b_{12}\\ 0& b_{22}\end{pmatrix}\) = \(\displaystyle \begin{pmatrix}
c_{11}& c_{12}\\ c_{21}& c_{22}\end{pmatrix}\)
and so forth. Thus we only need to do matrix multiplication like
\(\displaystyle A \begin{pmatrix}
x_{11}& 0\\ x_{21}& 0\end{pmatrix}\)
or
\(\displaystyle A \begin{pmatrix}
0& b_{12}\\ 0& b_{22}\end{pmatrix}\)
which can both be reduced to matrix multiplication like
\(\displaystyle A \begin{pmatrix}
x_{11}\\ x_{21}\end{pmatrix}\)
We now come to the idea of eigenvalues and eigenvectors. Without going into details, for which see the above link to start with, for each mXm square matrix there are m eigenvalues
λi,i=1,2,3,...,m and corresponding eigenvectors
vi,i=1,2,3,...,m such that
A v
i =
λi v
i
If the eigenvalues are all distinct and non-zero, the eigenvectors span the space. That is for any vector [in our 2X2 example space]
x =
(x11x21)
there exists coefficients a
1 and a
2 such that
x = a
1 v
1 + a
2 v
2.
Well now start multiplying by A
A x = A (a
1 v
1 + a
2 v
2) = a
1 Av
1 + a
2 Av
2 = a
1 λ1v
1 + a
2 λ2v
2
Multiply by A again
A
2 x = a
1 λ12v
1 + a
2 λ22v
2
...
A
n x = a
1 λ1nv
1 + a
2 λ2nv
2
Lets assume distinct eigenvalues with
∣λ1∣>∣λ2∣, then
A
n x =
λ1n[a1v1+a2(λ1λ2)nv2]
What happens to that
(λ1λ2)n
as n becomes large? What then happens to A
n x? What then happens to A
n?
EDIT: Fix grouping, i.e. I made a dumb mistake and fixed it.
