91Ó°ÊÓ

The transition matrix in Example 5 has the property that both its rows and its columns add up to 1 In general, a matrix \(A\) is said to be doubly stochastic if both \(A\) and \(A^{T}\) are stochastic. Let \(A\) be an \(n \times n\) doubly stochastic matrix whose eigenvalues satisfy \\[ \lambda_{1}=1 \quad \text { and } \quad\left|\lambda_{j}\right|<1 \text { for } j=2,3, \ldots, n \\] Show that if \(\mathbf{e}\) is the vector in \(\mathbb{R}^{n}\) whose entries are all equal to \(1,\) then the Markov chain will converge to the steady-state vector \(\mathbf{x}=\frac{1}{n} \mathbf{e}\) for any starting vector \(\mathbf{x}_{0} .\) Thus, for a doubly stochastic transition matrix, the steady-state vector will assign equal probabilities to all possible outcomes.

Let \(A\) be an \(n \times n\) matrix and let \(\lambda\) be a nonzero eigenvalue of \(A .\) Show that if \(\mathbf{x}\) is an eigenvector belonging to \(\lambda,\) then \(\mathbf{x}\) is in the column space of \(A .\) Hence the eigenspace corresponding to \(\lambda\) is a subspace of the column space of \(A\)

Let \(\lambda\) be an eigenvalue of an \(n \times n\) matrix \(A\) and let \(\mathbf{x}\) be an eigenvector belonging to \(\lambda\). Show that \(e^{\lambda}\) is an eigenvalue of \(e^{A}\) and \(\mathbf{x}\) is an eigenvector of \(e^{\mathbf{A}}\) belonging to \(e^{\lambda}\)