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\begin{document}

\title{Finding Eigenvalues and Eigenvectors}
\date{November 13, 2002}
\author{Blake Thornton}

\maketitle


\section{Finding eigenvalues and eigenvectors}

We start with a $n \times n$ matrix $A$.

\begin{enumerate}
\item  We want to solve the equation
  %
  \begin{equation} \label{ee one}
  A\mathbf{v} = \lambda \mathbf{v}
  \end{equation}
  %
  for both $\mathbf{v}$ and $\lambda$.

\item  It is easy to see that $\mathbf{v}=\mathbf{0}$ (the zero vector) 
  is always a solution to this equation
  (for any value of $\lambda$).
  We decide that this isn't too interesting and start looking for other solutions.

\item  We transform equation~(\ref{ee one}) into the equation:
  %
  \begin{equation} \label{ee two}
  (A-\lambda I)\mathbf{v} = \mathbf{0}
  \end{equation}
  This is now a bit more familiar.

\item  Previous results that we need to use:
  \begin{enumerate}
  \item  A matrix equation $M\mathbf{X}=\mathbf{0}$ always has at least one solution.
  \item  A matrix equation $M\mathbf{X}=\mathbf{0}$ always an infinite number of zeros if and only if
    one of the following equivalent conditions hold (why are all these equivalent?):
    \begin{enumerate}
    \item  When you row reduce $M$ and put $M$ into reduced row echelon form, 
      there is at least one row of zero's.
    \item  When you row reduce $M$ and put $M$ into reduced row echelon form, 
      there is at least one ``parameter column.''
    \item  Rank $M$ is less then $n$.
    \item  $\det M =0$.
    \end{enumerate}
  \end{enumerate}

\item  \textbf{Finding eigenvalues}\\
  Since we are trying to solve equation~(\ref{ee two}), 
  and we are looking for nontrivial solutions, we apply the results above.
  (What is a nontrivial solution?  What is a trivial solution?)
  There is a nontrivial solution if and only if
  \[
  \det(A-\lambda I) = 0
  \]
  Notice that this yeilds an equation only in $\lambda$, which we can solve.
  Here are some facts and definitions:
  \begin{enumerate}
  \item  $\det(A-\lambda I)$ is always a polynomial.  (Why?  What is the degree of this polynomial?)
  \item  $\det(A-\lambda I)$ is called the \emph{characteristic polynomial}.
  \item  $\det(A-\lambda I) =0$ is called the \emph{characteristic equation}.
  \item  Solving the characteristic equation give all the eigenvalues of the matrix $A$.
    (This does not give you any of the eigenvectors though!)
  \end{enumerate}

\item  \textbf{Finding eigenvectors}\\
  You now have all the eigenvalues for a matrix $A$.
  You now need to find the eigenvectors for each eigenvalue.
  So, start with an eigenvalue, $\lambda$.
  For a fixed $\lambda$, we need to solve equation~(\ref{ee two}):
  \[
  (A-\lambda I)\mathbf{v} = \mathbf{0}
  \]
  Or, in other words, we need to find the null space of the matrix $A-\lambda I$.
  In theory, we are already experts at doing this.

\end{enumerate}





\section{Properties of eigenvalues and eigenvectors}

Most of these are easy to prove 
(maybe I did it in class, it might be in the book or you can probably prove these statements yourself).
\\
(Homework: prove all these statements below)

\begin{enumerate}
\item  If $\mathbf{v}$ is an eigenvector of $A$ with eigenvalue $\lambda$, 
  then all scalar multiples of $\mathbf{v}$ are eigenvalues of $A$ with eigen value $\lambda$.

\item  Suppose that $\mathbf{v}$ and $\mathbf{u}$ are eigenvalues of $A$, 
  both with eigenvalue $\lambda$.
  Then, for any scalars, $a,b$, $a\mathbf{v}+b\mathbf{u}$ is an eigenvector with eigenvalue $\lambda$.

\item  For a given eigenvalue $\lambda$, the set of all eigenvectors with that eigenvalue
  is a subspace.  (The proof of this is basically the previous item)

  This is called the \emph{eigenspace} corresponding to the eigenvalue.

\end{enumerate}


\newpage
\section{Examples}

Hopefully, we'll work these out in class.
Some of these are right from the book.
(You should try to work them out in Maple to save work!)
For these examples we will:
\begin{enumerate}[(a)]
\item  Find all eigenvalues of the matrix:
  \begin{enumerate}
  \item  Find the characteristic polynomial and characteristic equation.
  \item  Solve the characteristic equation.
  \end{enumerate}
\item  Find a basis for each eigenspace.
  (basically this means we will find all eigenvectors).
\item  Maybe do an application of power of matrices similar to what is done in the book.
\end{enumerate}

\begin{enumerate}

\item  
  \[
  A = \left[
    \begin{array}{rr}
      3 & 1 \\
      1 & 3
    \end{array}
    \right]
  \]

\item  
  \[
  A = \left[
    \begin{array}{rrr}
      3 & 1 & 0 \\
      0 & 1 & 0 \\
      4 & 2 & 1
    \end{array}
    \right]
  \]

\item  
  \[
  A = \left[
    \begin{array}{rrr}
      3 & 1 & 1 \\
      0 & 3 & 1 \\
      0 & 0 & 5
    \end{array}
    \right]
  \]

\item  
  \[
  A = \left[
    \begin{array}{rrr}
      1 & 2 & 3 \\
      0 & 5 & 6 \\
      6 & 1 & 10
    \end{array}
    \right]
  \]

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



\end{enumerate}



\end{document}






