Proving a specific eigenvalue of a 6x6 Matrix.

Question

Currently, I am looking at Markov diagrams and their associated transition matrices. I am trying to prove that the transition matrix has eigenvalue $\lambda=1$. I am aware that to find the eigenvalues of a matrix we use:

$$\det (A - \lambda I_{6}) = 0$$

and then each $\lambda$ gives an eigenvalue of the matrix. But this seems like a very lengthy process for a $6 \times 6$ matrix.

I'm also aware that every stochastic matrix has the eigenvalue of $\lambda=1$ but I would like to learn how to work this out for this specific matrix.

$$ \begin{bmatrix} 0 & \frac{1}{2} & \frac{1}{3} & 0 & 0 & 0 \\ \frac{1}{2} & 0 & 0 & \frac{1}{3} & 0 & 0 \\ \frac{1}{2} & 0 & 0 & \frac{1}{3}& \frac{1}{2}& 0 \\ 0 & \frac{1}{2} & \frac{1}{3} & 0 & 0 & \frac{1}{2} \\ 0 & 0 & \frac{1}{3}& 0 & 0 & \frac{1}{2} \\ 0 & 0 & 0 & \frac{1}{3} & \frac{1}{2} & 0 \\ \end{bmatrix} $$

Any help would be greately appreciated.

edit

Having using the eigenvector of $\large[1\ 1\ 1\ 1\ 1\ 1\ \large]$ to show that $\lambda =1$ is an eigenvalue I now want to find a stable distribution.

So I am going to labe my horizontal and vertical entries M, E, C, S, W, N respectively. and will set $\large[M\ E\ C\ S\ W\ N \large]$ to be a vector to multiply my transition matrix by.

So i get:

$1/2E+1/3C = M$

$1/2M+1/3S=E$

$1/2M+1/3S+1/2W=C$

$1/2E+1/3C+1/2N=S$

$1/3C+1/2N=W$

$1/3S+1/2W=N$

So Im wondering if what I'm doing is right and whether my next steps should be to try and work out what the values of M E C S W N are?

Answer 1

HINT

If you are looking for a specific eigenvalue, compute the matrix $B = A - \lambda I$, and show that $\det(B) = 0$.

If you can guess the corresponding eigenvector $\vec{x}$, it becomes even easier. Multiply matrices to show that $$ A \vec{x} = \lambda \vec{x} $$ and you are done.

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In your case, however, it is even easier. The sum of all entries in each column is $1$, which makes you matrix stochastic. For such a matrix, the vector of all $1$'s is a left eigenvector, in other words, $$ \begin{bmatrix} 1 & 1 & 1 & 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} 0 & \frac{1}{2} & \frac{1}{3} & 0 & 0 & 0 \\ \frac{1}{2} & 0 & 0 & \frac{1}{3} & 0 & 0 \\ \frac{1}{2} & 0 & 0 & \frac{1}{3}& \frac{1}{2}& 0 \\ 0 & \frac{1}{2} & \frac{1}{3} & 0 & 0 & \frac{1}{2} \\ 0 & 0 & \frac{1}{3}& 0 & 0 & \frac{1}{2} \\ 0 & 0 & 0 & \frac{1}{3} & \frac{1}{2} & 0 \\ \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 & 1 & 1 & 1 \end{bmatrix} $$

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UPDATE

Typically to find the eigenvectors and eigenvalues of a matrix $A$, first solve $\det(A-\lambda I)=0$ and then when you get the eigenvalues, plug them into $(A - \lambda I) \vec{x}=\vec{0}$ and solve for each $\vec{x}$ separately.

Answer 2

In order to prove that 1 is indeed an eigenvalue, you just need to prove that $\det(A-I_6)=0$. You can prove that by finding a linear combination of the rows or columns that are $0$.

Here, you can use that the matrix is stochastic, so the sum of each row is 1, when you substract $I_6$, you get that the sum of all the rows of $A-I_6 $ is 0, so 1 is an eigenvalue of A