Finding an eigenvector for a matrix such that the entries of each column sum 1

Question

Let $A \in \mathbb{R}^{n \times n}$ a matrix such that each entrie of $A$ (I denoted them by $\left[ A \right] _{ij}$) satisfies that $\left[ A \right] _{ij} \geq 0$ and if you sum the entries of each column, the result is $1$ (i.e. $\left[ A \right] _{1j}+...+\left[ A \right] _{nj}=1$ for all $j \in \{ 1, ... , n \}$). I need to find an explicit eigenvector of $A$ such that $Ax=x$. How can I find one? (it could be by inspection, the problem says: "guess one eigenvector such that...") I tried a lot, but unsuccesfully.

Answer 1

The fact that the entries in each column add to 1 means that the matrix $A-I$ is non-singular. Therefore the columns of $A-I$ are linearly dependent. If you can spot the dependency relationship then that gives you the eigenvector.

In general, the solution is a solution of the equations given by $(A-I)V=0$.

Are you sure the source of this problem is not putting eigenvectors on the left?

The reason the problem is not a reasonable one to ask is because the first $n-1$ rows of $A$ are 'arbitrary' - then choose the final row so all column sums are one. So basically you are being asked to give a solution of a general set of equations.

Answer 2

Your $A$ is column-stochastic. If it is also positive, or in general, if it is primitive, then $x$ is the steady-state distribution associated with the Markov chain defined by $A$, i.e. $\lim_{k\to\infty}A^k=xe^T$, where $e=(1,\ldots,1)^T$. In turn, $$ x=\lim_{k\to\infty}A^kp $$ where $p$ is any vector whose entries sum to $1$, such as $p=e_1=(1,0,\ldots,0)^T$. In particular, every column of $A^k$ converges to $x$ when $k$ tends to infinity.