Question

Alice is visiting a website with n pages. The structure of the website and Alice's browsing process is captured by a matrix A as follows: every hour if Alice is browsing the page i, she will goes to the page j next hour with probability Aj (assume that each entry of A is non-negative, and the entries in each row add up to 1) (a) Show that λ is an eigenvalue of A. (b) If Alice starts browsing from page i, show that after t hours, the probability that she ends up browsing page j is (A )ij.

Answer 1

a) Suppose Alice starts with page ; then, in the next hour she will be at exactly one of the pages, $j=1,\cdots,n$ with probabilities $A_{i1},\cdots,A_{in}$ respectively. Thus, the probability that Alice will be at any one of the pages, $j=1,\cdots,n$ is

41 +...+ Ain

But this probability is because in the next hour she will definitely be at one of the pages; this means

41 +...+ Ain-1

and this is true for all $i=1,\cdots,n$. Let

$v=\begin{pmatrix}1\\ \cdot\\\cdot\\ 1\end{pmatrix}$

Then

$(Av)_i=\sum_{j=1}^nA_{ij}v_j=\sum_{j=1}^nA_{ij}=1=v_i$

for all $i=1,\cdots,n$. This means $Av=v$. Thus, $\lambda=1$ is an eigenvalue of .

b) We use induction on . If $t=1$ then by definition the probability that Alice ends up browsing page (starting from page ) after hours is $A_{ij}=(A^t)_{ij}$ . Thus, the statement in question is true for $t=1$.

Suppose that the statement is true for some
t >1
. Now suppose that Alice starts at page . Suppose that after hours she is at page . Then the probability that she will be at page in hour is (by base case and induction hypothesis)

$(A^t)_{ik}A_{kj}$

Since can be any of the pages $k=1,\cdots,n$, the probability that she is at page in hour is

$\sum_{k=1}^n(A^t)_{ik}A_{kj}=(A^tA)_{ij}=(A^{t+1})_{ij}$

Thus, the statement in question is true for . By induction, the statement is true for all
t >1
​​​​​​​.