Problem involving Markov matrices. The question is:

Cecil Sagehen is either happy or sad. If he is happy one day, then he is happy the next day three times out of four. If he is sad one day, then he is sad the next day one time out of three. During the coming year, how many days do you expect Cecil to be happy?

I construct the transition matrix corresponding to this problem,
$$M = \begin{pmatrix} \frac{3}{4} & \frac{1}{4} \\
\frac{2}{3}&  \frac{1}{3} 
\end{pmatrix} $$
We have that the eigen values of $M$ are $1$ and $\frac{1}{12}$ with corresponding eigen vectors $v_1 = (1,1)$ and $v_2 = (-3,8)$. I know that we can diagnolize $M$ and then compute $M^{365}$, but is there a way to solve this problem without diagonalizing; as in, can we solve this problem just using the eigen vectors $v_1$ and $v_2$? Any help would be appreciated. Thanks!
 A: I strongly suspect that you are meant to use the following approximation of the result: the final distribution of the Markov chain is the eigenvector of $M^T$ associated with the eigenvalue $1$, with non-negative entries normalized to have sum $1$. In this case, we find that eigenvector to be $(8/11,3/11)$. Correspondingly, we find that after a sufficiently long time amount of time, the probability of being happy at that moment is given by $8/11$.
Correspondingly, we should expect that out of $365$ days, Cecil should be happy on $365 \cdot 8/11 = 265.45 \approx 265$ days.

If we were to compute the exact result, then given an initial distribution $\pi = (\pi_1,\pi_2)^T$, we would need to compute
$$
\pi^T[I + M + M^2 + \cdots + M^{364}].
$$
Using your eigenvalues, we have $M = PDP^{-1}$ where
$$
P = \pmatrix{1&3\\1&-8} \implies P^{-1} = \frac 1{11} \pmatrix{8&3\\1&-1}, \quad D = \pmatrix{1&0\\0& \frac 1{12}}.
$$
We note that for $r \neq 1$, we have $1 + r + \cdots + r^{364} = \frac{1-r^{365}}{1-r}$, and for $r = 1$ this sum is equal to $365$. Putting all that together, we have
$$
\pi^T[I + M + M^2 + \cdots + M^{364}] = \\
\pi^T P [I + D + D^2 + \cdots + D^{364}]P^{-1}\pmatrix{1\\0}  = \\
\pi^T P \pmatrix{365 & 0\\0 & \frac{1 - (1/12)^{364}}{1-(1/12)}}P^{-1}\pmatrix{1\\0} \approx\\
\pi^T P \pmatrix{365 & 0\\0 & \frac{1}{1-(1/12)}}P^{-1}\pmatrix{1\\0} =\\
 \pi^T P \pmatrix{365 & 0\\0 & \frac{12}{11}}P^{-1}\pmatrix{1\\0} =\\
\frac 1{11}\pi^T\pmatrix{1&3\\1&-8} \pmatrix{365&0\\0&\frac{12}{11}}\pmatrix{8&3\\1&-1}\pmatrix{1\\0} =\\
\pi^T\left(\begin{matrix}265.75 & 99.25\\264.66 & 100.34\end{matrix}\right) \pmatrix{1\\0}
\\
265.75\,\pi_1 + 264.66\,\pi_2
$$
If we start with the initial distribution $(1/2,1/2)$ (i.e. Cecil is equally likely to be happy or unhappy on the first day), this leads to the answer $265.21 \approx 265$. So, it indeed seems that the final distribution yields a close approximation to the true answer.
Based on the coefficients of the answer given above, we can see that whatever the initial distribution, the answer will lie within the interval $[264.66, 265.75]$.
