Proving a specific eigenvalue of a 6x6 Matrix. 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?
 A: 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.

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

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.
A: 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
