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\textbf{Exercises on diagonalization and powers of A}
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\begin{problem} (6.2 \#6.  {\em Introduction to Linear Algebra:} Strang) Describe all matrices $S$ that diagonalize this matrix $A$ (find all eigenvectors):

$$ A = \twobytwo 4 0 1 2.$$

\noindent Then describe all matrices that diagonalize $A^{-1}.$
\end{problem}

\solution{ To find the eigenvectors of $A,$ we first find the eigenvalues:
$$\det \twobytwo {4-\lambda} 0 1 {2-\lambda} = 0 \Longrightarrow
  (4-\lambda)(2-\lambda)=0.$$  Hence the eigenvalues are $\lambda_1 =
  4$ and $ \lambda_2 =2.$ Using these values, we find the eigenvectors
  by solving $(A - \lambda I)\vb x = \vb 0:$
$$(A - \lambda_1 I)\vb x = \twobytwo 0 0 1 {-2} \twobyone {y} {z} = \twobyone 0 0 \Longrightarrow y = 2z,$$
thus any multiple of \textbf{(2,1)} is an eigenvector for $\lambda_1.$
$$(A - \lambda_2 I)\vb x = \twobytwo 2 0 1 0 \twobyone y z = \twobyone 0 0 \Longrightarrow y=0, z=\text{free variable,} $$
thus any multiple of \textbf{(0,1)} is an eigenvector for $\lambda_2. $ Therefore the columns of the matrices $S$ that diagonalize $A$ are nonzero multiples of (2,1) and (1,0).  They can appear in either order. 

Finally, because $A^{-1} = S \Lambda^{-1}S^{-1}$ the same matrices $S$ will diagonalize  $A^{-1}$.
}

\begin{problem} (6.2 \#16.)  Find $\Lambda$ and $S$ to diagonalize $A:$
$$A = \twobytwo {.6} {.9} {.4} {.1} . $$
What is the limit of $\Lambda^k$ as $k \rightarrow \infty$? What is the limit matrix of $S \Lambda^k S^{-1}$?  In the columns of this matrix you see the \rule{20mm}{0.4pt}.
\end{problem}

\solution{ Since each of the columns of $A$ sum to one, $A$ is a Markov matrix and definitely has eigenvalue $\lambda_1 =1.$ The trace of $A$ is $.7$, so the other eigenvalue is $\lambda_2 = .7 -1 = -.3.$ To find $S$ we need to find the corresponding eigenvectors:
$$(A-\lambda_1 I)\vb {x_1}= \twobytwo {-.4} {.9} {.4} {-.9} \twobyone
  y z = \twobyone 0 0 \Longrightarrow \vb {x_1} = (9,4).  $$
$$(A-\lambda_2 I)\vb {x_2}= \twobytwo {.9} {.9} {.4} {.4} \twobyone y z = \twobyone 0 0 \Longrightarrow y = -z \Longrightarrow  \vb {x_2} = (1,-1).$$
Putting these together, we have:
$$S = \twobytwo 9 1 4 {-1} \text{ and }  \Lambda = \twobytwo 1 {} {} {-.3} \text{.  As } k\rightarrow \infty, \hspace{6pt} \Lambda^k \rightarrow \twobytwo 1 {} {} 0.$$
So
$$S \Lambda^k S^{-1} \rightarrow \twobytwo 9 {1} 4 {-1} \twobytwo 1 {} {} 0 \left(\frac{1}{13} \right) \twobytwo 9 1 4 {-9} = \frac{1}{13} \twobytwo 9 9 4 4.$$
In the columns of this matrix you see the \textbf{steady state vector.}
}

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