Let $ A $ be a $ 4 \times 4 $ matrix such that the sum of elements in each row is $ 1 $. Find out the sum of all the elements in $ A ^ { 10 } $.
I don't know anything about how to get a connection between the two statements without actually calculating $ A ^ { 10 } $, which will not be an efficient process.
Also, I don't know about eigenvalues and eigenvectors which were used in some questions I saw for determining the relationship between $ A $ and $ A ^ { – 1 } $. I don't think the inverse calculation is useful in this question.
What should be understood to solve this?
Best Answer
HINT Before we see what happens in $A^{10}$, let's explore $A^2$.
So let $A = (a_{ik})_{i,k=1}^n$ with $\sum_{k=1}^n a_{ik} = 1$ for all $i \in [n] = \{1, 2, \ldots, n\}$.
Note that if $A^2 = (b_{ik})_{i,k=1}^n$ then $$ b_{ik} = \sum_{j=1}^n a_{ij} a_{jk} $$ and the sum of one row would be $$ \sum_{i=1}^n b_{ik} = \sum_{i=1}^n \sum_{j=1}^n a_{ij} a_{jk} = \sum_{j=1}^n a_{jk} \left(\sum_{i=1}^n a_{ij}\right) = \sum_{j=1}^n a_{jk} \cdot 1 = 1. $$ Can you follow this logic to see what happens in $A^{10}$?
UPDATE Your supposition that every row of $A^k$ will sum to $1$ for all positive integers $k$ is correct. One way to convince yourself of that is that every stochastic matrix (one where each row sums to $1$) can be seen as a transition matrix of a Markov Chain with states $[n]$. Then, $A^2$ is the matrix of 2-step transitions, and $A^k$ is the matrix of $k$-step transitions, both of which must be stochastic as well...