Dear Michelle,

thank you for the matrix. Here are some comments. I first save the matrix

matrix = {{0, 1, 0, 0, 0, 0, 1, 0}, {0, 0, 0, 0, 0, 1, 1, 0}, {0, 1,

0, 0, 1, 0, 0, 0}, {1, 0, 1, 0, 0, 0, 1, 1}, {0, 0, 0, 0, 0, 1, 0,

1}, {1, 0, 0, 0, 0, 0, 0, 1}, {0, 0, 0, 1, 1, 1, 0, 0}, {0, 1, 0,

0, 0, 0, 1, 0}}

I then plot it.

AdjacencyGraph[matrix, VertexLabels -> "Name", ImagePadding -> 10]

Looks nice:

Next I calculate the eigenvalues - strictly speaking this is not necessary for your problem.

Eigenvalues[matrix] // N

This gives

{2.3728, -0.680664 + 1.64275 I, -0.680664 - 1.64275 I, -1.23221,

0.295924 + 0.70383 I, 0.295924 - 0.70383 I, -0.371109, 0.}

The normalised eigenvector to the largest (first) eigenvalue is obtained by

Normalize[Part[Eigenvectors[matrix] // N, 1]]

Now that gives the desired result

{0.31902, 0.304159, 0.232608, 0.557762, 0.247773, 0.268897, 0.452812, 0.31902}

which are the relative times to spend at the different nodes.

If we normalise the sum to 1, we obtain the percentage of the time that a random walker would spend at a given node.

Part[Eigenvectors[matrix] // N, 1] / Total[Part[Eigenvectors[matrix] // N, 1]]

This gives

{0.118066, 0.112566, 0.0860857, 0.206422, 0.0916982, 0.0995159, 0.167581, 0.118066}

So about 12% of the time the random walker is at node one etc.

Just to check the result, let's use the power algorithm to determine the eigenvector. I define an arbitrary starting vector:

vecstart = Table[1., {i, 1, Length[matrix]}]

(*{1., 1., 1., 1., 1., 1., 1., 1.}*)

I then apply the matrix to the vector and iterate...say 100 times:

Nest[Normalize[matrix.#] & , vecstart, 100]

The result is

{0.31902, 0.304159, 0.232608, 0.557762, 0.247773, 0.268897, 0.452812, 0.31902}

which is identical to the one above. Normalising this to 1 in the numerically most clumsy way

Nest[Normalize[matrix.#] & , vecstart, 100]/Total[Nest[Normalize[matrix.#] & , vecstart, 100]]

gives the relative times spent at the nodes

{0.118066, 0.112566, 0.0860857, 0.206422, 0.0916982, 0.0995159, 0.167581, 0.118066}

I hope that this helps. Note that it does not take the probabilities that you gave into consideration.

M.