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\begin{document}

\title{Diagonalization}
\author{Blake Thornton}
\date{November 20, 2002}

\maketitle



\section{Definitions and Theorem}

\begin{defn}
An $n\times n$ matrix $A$ is \emph{diagonalizable}
 if there is some matrix $Q$ such that $Q^{-1}AQ$ is diagonal.
\end{defn}


\begin{theorem} \label{n evecs}
An $n \times n$ matrix $A$ is diagonalizable if and only if $A$ has a set of $n$
linearly independent eigenvectors.
\end{theorem}
What this theorem means is that if you can find a basis for $\R^n$, 
consisting of eigenvectors, then you can diagonalize the matrix.

\begin{defn}
Let $A$ be a $n \times n$ matrix and let $\lambda_1, \ldots, \lambda_k$ be all the different eigenvalues
of $A$.
Let $n_1, \ldots, n_k$ be the multiplicities of the eigenvalues 
(so $n_i$ is the multiplicity of the eigenvalue $\lambda_i$).
Then $A$ is \emph{deficient} if for any of the eigenvalues, 
the eigenspace associated to the eigenvector $\lambda_i$ has dimension less then $n_i$.
\end{defn}
Basically what this definition is saying is that you should be able to ``see'' the dimension
of the eigenspace by looking at the characteristic polynomial and looking at the 
multiplicity of the roots.
If the eigenspace turns out to have not big enough dimension, then the matrix is deficient.

\begin{theorem} \label{not deficient}
A matrix is diagonalizable if and only if it is not deficient.
\end{theorem}

\begin{theorem} \label{n different}
Let $A$ be a $n \times n$ matrix with eigenvalues $\lambda_1, \ldots, \lambda_k$.
Let $Q_1, \ldots, Q_k$ be eigenvectors associated to the eigenvalues.
Then $\{ Q_1, \ldots, Q_k \}$ are independent.
\end{theorem}
What this is saying is that different eigenvalues give different eigenvectors.
Or, more precisely, different eigenvectors give linearly independent eigenvectors.





\section{Procedure}

Here I will outline how to go about diagonalizing a matrix.
There are a couple of points that need addressed.
Here are the main issues that you want to address:
\begin{enumerate}[I.]
\item  Can you diagonalize the matrix?
\item  What will the diagonalized matrix look like? (What is $D$?)
\item  What is the diagonalizing matrix? (What is the matrix $Q$?)
\end{enumerate}

Here is the procedure.
You should notice that these are the steps that I followed in class and 
also the steps that the book follows (although probably not as clearly laid out for you).
We will assume that $A$ is a $n\times n$ matrix.

\begin{enumerate}

\item  Find all the eigenvalues of $A$.
  \\
  When you do this, you should notice:
  \begin{enumerate}
  \item  What are the multiplicities of each eigenvalues.
    (You should be able to see this if you factored the characteristic polynomial.)
    If you can't see this, it is not the end of the world, but this will tell you want dimension you 
    are expecting each eigenspace to be.
  \end{enumerate}

\item  Find a basis for each eigenspace.  
  \\
  When doing this notice:
  \begin{enumerate}
  \item  Is the matrix deficient?  
    You should be able to see this by either comparing the dimension of each eigenspace to 
    the multiplicity of eigenvector found above.
    Otherwise, you can just see if you found $n$ independent eigenvectors.
  \end{enumerate}

\item  Is the matrix diagonalizable?
  \\
  Ways to answer this question:
  \begin{enumerate}
  \item  Are there $n$ different eigenvalues?  
    If so, then by Theorem~\ref{n different} $A$ is diagonalizable.
  \item  Is $A$ deficient?
    If so, then by Theorem~\ref{not deficient} $A$ is not diagonalizable.
    If not, then $A$ is diagonalizable.
  \item  Is there a basis for $\R^n$ consisting of eigenvectors?
    If so, then by Theorem~\ref{n evecs} $A$ is diagonalizable.
    If not, then $A$ is not diagonalizable.
  \end{enumerate}

\item  Find the diagonalizing matrix $Q$:
  \begin{enumerate}
  \item  Let $Q$ be the $n \times n$ matrix with the $n$ linearly independent eigenvectors as columns.
  \end{enumerate}

\item  Find the diagonal matrix $D$.
  \begin{enumerate}
  \item  Then $D=Q^{-1}AQ$ is the diagonal matrix.
    $D$ should be the matrix with eigenvalues on the diagonal 
    (in the same order that you put the eigenvectors into the matrix $Q$).
  \item  I find it confusing to remember if $D=Q^{-1}AQ$ or if $D=QAQ^{-1}$.
    I think the easiest way to remember this is the remember the chart I drew on the board
    on November 18 and to remember what the matrix $Q$ does.
    \\
    Let $S$ be the standard basis in $\R^n$ and let $E$ be the basis of eigenvectors in $\R^n$.
    Let $(\R^n,S)$ be $\R^n$ written in $S-coordinates$ and $(R^n,E)$ be $\R^n$ in $E$-coordinates:
    \[
    \begin{CD}
      (\R^n, S) @> {A} >> (\R^n, S) \\
      {Q}@AAA   Q@AAA \\
      (\R^n,E) @> {D} >> (\R^n, E) \\
    \end{CD}
    \]
  \end{enumerate}

\end{enumerate}



\section{Exercises}

Diagonalize the following matrices.
This means that you must do the following:
\begin{enumerate}[(a)]
\item  Find all eigenvectors and eigenvalues.
\item  Determine if the matrix is diagonalizable.
\item  Find the diagonalizing matrix $Q$.
\item  Find the diagonal matrix $D$.
\item  (Using the results from matrix exponentials, find the exponential of each matrix too)
\end{enumerate}


\begin{enumerate}
\item  
  \[
  \left[
    \begin{array}{rr}
      1 & 1 \\
      1 & 1
    \end{array}
    \right]
  \]

\item  
  \[
  \left[
    \begin{array}{rrr}
      1 & 0 & 0 \\
      -1 & 3 & 0 \\
      3 & 2 & -2
    \end{array}
    \right]
  \]

\item  
  \[
  \left[
    \begin{array}{rrr}
      3 & -11 & 16 \\
      2 & -8 & 8 \\
      1 & -3 & 2
    \end{array}
    \right]
  \]

\item  
  \[
  \left[
    \begin{array}{rrr}
      2 & 3 & 3 \\
      3 & 2 & 3 \\
      3 & 3 & 2
    \end{array}
    \right]
  \]

\item  
  \[
  \left[
    \begin{array}{rrrr}
      3 & -1 & 0 & 2\\
      0 & 3 & 1 & 0 \\
      0 & 0 & -1 & 2\\
      0 & 0 & 1 & 0
    \end{array}
    \right]
  \]

\item  
  \[
  \left[
    \begin{array}{rrr}
      1 & -1 & 4 \\
      3 & 2 & -1 \\
      2 & 1 & -1
    \end{array}
    \right]
  \]

\end{enumerate}



\section{Application - Matrix exponentials}

I am not going to motivate this too deeply by giving lots of reasons to care about
matrix exponentials.
You are welcome to come talk to me about it, but matrix exponentials are often used in:
\begin{itemize}
\item  Differential equations.
\item  Geometry (this is where I have used them)
\end{itemize}

The basis point is that given a matrix $A$, compute $\exp(A)$ where 
\begin{eqnarray*}
\exp(A) & = & I + A + \frac{1}{2!}A^2 + \frac{1}{3!}A^3 + \cdots \\
  & = & \sum_{n=0}^\infty \frac{1}{n!}A^n
\end{eqnarray*}

Here is how to compute $\exp(A)$ (there are some gaps in this process that you need to fill in):
\begin{enumerate}
\item  Diagonalize $A$ (if possible).
\item  Write $A=QDQ^{-1}$ and plug into the formula for $\exp(A)$:
  \begin{eqnarray*}
    \exp(A) & = & \sum_{n=0}^\infty \frac{1}{n!}A^n \\
       & = & \sum_{n=0}^\infty \frac{1}{n!}(QDQ^{-1})^n \\
       & = & Q \left(\sum_{n=0}^\infty \frac{1}{n!}D^n \right)Q^{-1} \\
       & = & Q(\exp(D))Q^{-1}
  \end{eqnarray*}
\item  But, $\exp(D)$ should be easy since $D$ is diagonal -- the exponential of a diagonal 
  is the diagonal matrix made up of the exponential of the diagonal elements of $D$.
\end{enumerate}

\subsection{Example}

\[
A = \left[
  \begin{array}{rr}
    3 & 1 \\
    1 & 3
  \end{array}
  \right]
\quad
D = \left[
  \begin{array}{rr}
    2 & 0 \\
    0 & 4
  \end{array}
  \right]
\quad
Q = \left[
  \begin{array}{rr}
    -1 & 1 \\
    1 & 1
  \end{array}
  \right]
\]
\[
\exp(D) = \left[
  \begin{array}{rr}
    e^2 & 0 \\
    0 & e^4
  \end{array}
  \right]
\]
\begin{eqnarray*}
\exp(A) & = & Q\exp(D) Q^{-1} \\
   & = & 
   \left[
     \begin{array}{rr}
       \frac{1}{2}(e^4+e^2) & \frac{1}{2}(e^4-e^2) \\
       \frac{1}{2}(e^4-e^2) & \frac{1}{2}(e^4+e^2)
     \end{array}
     \right]
\end{eqnarray*}



\end{document}






