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Exercise 5.3.19

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1.

First we check that

{(cM + (1 - c) N)}^{t} u = c M^{t} u + (1 - c) N^{t} u = cu + (1 - c) u = u .

Thus the new matrix is a transition matrix by the Corollary after Theorem 5.15. So it’s enough to show that the new matrix is regular. Now suppose that $M^{k}$ is positive. Then we know that ${(cM + (1 - c) N)}^{k}$ is the sum of $c^{k} M^{k}$ and some lower order terms, which are nonnegative. So we know that it’s a positive matrix and so the new matrix is regular.

2.

Pick a scalar

d

such that each entry of

d M^{'}

is larger than that of

M

. Then we may pick

c = \frac{1}{d}

and

N = \frac{1}{1 - c} (M^{'} - cM)

and know that $N$ is nonnegative. Finally we may check that

N^{t} u = \frac{1}{1 - c} ({M^{'}}^{t} u - c M^{t} u) = \frac{1}{1 - c} \cdot (1 - c) u = u .

So $N$ is also a transition matrix by the Corollary after Theorem 5.15.

3.

By symmetry, it’s enough to prove only the one side of that statement. Also, by (b) we could write

M^{'} = cM + (1 - c) N

for some scalar $c$ and some transition matrix $N$ . Now, if $M$ is regular, then $M^{'}$ is also regular by (a).

jephianlin

2011-06-27 00:00

Exercise 5.3.19

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