Eigenvalue & Eigenvector question So far, I have part (A) done of this question, but I am REALLY stumped at part b and c. Can anyone perhaps help me out?
The city of Mawtookit maintains a constant population 300,000 people from year to year. A political science study estimated that there were 150,000 Independents,90,000 Democrats, and 60,000 Republicans in the town. It was also estimated that each year 20 percent of the Independents become Democrats and 10 percent become Republicans. Similarly, 20 percent of the Democrats become Independents and 10 percent become Republicans, while 10 percent of the Republicans defect to the Democrats and 10 percent become Independents each year. Let
x = [150,000 , 90,000 , 60,000]
and let x(1) be a vector representing the number of people in each group after 1 year.
a) Find a matrix $A$ such that $A$*x*= x(1)
b) Show that $λ_1=1.0  λ_2=0.5$  and $λ_3=0.7$ are eigenvalues of $A$, and factor $A$ into a product $XDX-I$, where $D$ is diagonal.
c) Which group will dominate in the long run? Justify your answer by computing $\lim_{n\to \infty}A^{(n)}$*x*
For part a I managed to develop a matrix that well list the number of Independents, Republicans, and Democrats in one year. the matrix I have goes as follows:
$A =\begin{pmatrix}
0.7&0.2&0.1\\0.2&0.7&0.1\\0.1&0.1&0.8
\end{pmatrix}$
not sure what to do for parts b & c though. 
 A: Hints:
To find the eigenvalues of: $$A =\begin{pmatrix}
0.7&0.2&0.1 \\ 0.2&0.7&0.1 \\ 0.1&0.1&0.8
\end{pmatrix},$$  
we setup $|A - \lambda I| = 0$ and solve the characteristic polynomial, so we have:
$$|A -\lambda I| = \begin{pmatrix}
0.7 - \lambda & 0.2 & 0.1\\0.2 & 0.7 -\lambda & 0.1\\0.1 & 0.1 & 0.8 - \lambda
\end{pmatrix} = 0$$  
From this, we get the characteristic polynomial as: $$-\lambda^3+ 2.2 \lambda^2 -1.55 \lambda + 0.35 = 0$$
This gives us three eigenvalues: $λ_1=1.0, λ_2=0.5$  and $λ_3=0.7$
Now, how do you factor $A$ into a product $XDX-I$, where $D$ is a diagonal?
Also, once you have a diagonal matrix, how can use that in order to find $\lim_{n\to \infty}A^{(n)}$?
A: We have $A = \frac{1}{10}\begin{bmatrix} 7&2&1\\2&7&1\\1&1&8 \end{bmatrix}$, and note that $\det (\lambda I -10A) =\lambda^3-22 \lambda^2+155 \lambda -350$, which factors into $(\lambda-10)(\lambda-5)(\lambda-7)$. Hence the eigenvalues of $A$ (not $10A$!) are $\lambda_1 = 1, \lambda_2 =\frac{1}{2}, \lambda_3=\frac{7}{10} $.
Grinding through the computations gives corresponding eigenvectors of $u_1 = \begin{bmatrix} 1 \\ 1 \\ 1\end{bmatrix}$, $u_2 = \begin{bmatrix} 1 \\ -1 \\ 0\end{bmatrix}$ and $u_3 = \begin{bmatrix} 1 \\ 1 \\ -2\end{bmatrix}$. Let $U =\begin{bmatrix} u_1 & u_2 & u_3\end{bmatrix}$, $\Lambda =\begin{bmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{2} & 0 \\ 0 & 0 & \frac{7}{10}\end{bmatrix}$. Then we have $A = U \Lambda U^{-1}$.
Presumably you meant $X D X^{-1} -I$, so letting $X = U$, and $D = \Lambda + I$ gives the required factorization (although it is not clear to me how this factorization helps).
To see which group dominates, we notice that $A^n = U \Lambda^n U^{-1}$, and it should be clear that $\lim_{n \to \infty} \Lambda^n = \hat{\Lambda} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$, hence $\lim_{n \to \infty} A^n =U \hat{\Lambda} U^{-1} = \frac{1}{3}\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$. This gives $\lim_{n \to \infty} A^n x = \begin{bmatrix} 100000 \\ 100000 \\ 100000 \end{bmatrix}$.
