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\title{Sturm Liouville operators vs symmetric matrices} 
\author{Eli, APM105b}
\maketitle

Functions may be thought of as (infinite-dimensional) vectors.  just
sample the function $f(x)$ at a set of values $x$ given by $x_n$ to
get $f_n=f(x_n)$, and think of $\{f_1,f_2,...,f_N\}$ as an
$N$-dimensional approximate vector representation of the function
$f(x)$.  At the limit of a full resolution of $f$ using an infinite
number of $x$ points, this becomes an infinite dimensional vector.

This justifies looking at vectors and matrices as parallels of
functions.  In particular, suppose that a matrix $A$ satisfies $A^T=A$
so that it is symmetric, and let its eigenvalues and vectors be given
by $A\hat{\mathbf{e}}_i=\lambda_i\hat{\mathbf{e}}_i$.  Then,
\begin{enumerate}
\item $\lambda_i$ are real;
\item eigenvectors belonging to different eigenvalues are orthogonal
  $\hat{\mathbf{e}}_i\cdot\hat{\mathbf{e}}_j=\delta_{ij}$;
\item eigenvectors are a complete set, for any vector $\mathbf{x}$,
  can write
  \begin{equation*}
    \mathbf{x}={\sum_n}a_n\hat{\mathbf{e}}_n={\sum_n}
    \frac{\mathbf{x}\cdot\hat{\mathbf{e}}_n}{\hat{\mathbf{e}}_n\cdot\hat{\mathbf{e}}_n}
    \hat{\mathbf{e}}_n
  \end{equation*}
  where
  $a_n=\mathbf{x}\cdot\hat{\mathbf{e}}_n/(\hat{\mathbf{e}}_n\cdot\hat{\mathbf{e}}_n)$
  and we can normally assume that the eigenvectors are normalized such
  that
  $\hat{\mathbf{e}}_n\cdot\hat{\mathbf{e}}_n=|\hat{\mathbf{e}}_n|^2=1$;
\end{enumerate}
This is all very similar to the S-L theorem for operators and
functions.

\bigskip

Why the similarity to symmetric matrices rather than some other type
of matrices?  It turns out that Self-adjoint operators are like
symmetric matrices.  To see this, consider that we may write scalar
multiplication of two vectors as
\begin{equation*}
  \mathbf{x}\cdot\mathbf{y}={\langle}\mathbf{x},\mathbf{y}{\rangle}
  =\mathbf{x}^T\mathbf{y},
\end{equation*}
where $\mathbf{x}^T$ is the transpose of $\mathbf{x}$.  Now, if $L$ is
a symmetric matrix, $L=L^T$ we can write
\begin{equation*}
  {\langle}L\mathbf{x},\mathbf{y}{\rangle}
  =(L\mathbf{x})^T\mathbf{y}
  =(\mathbf{x}^TL^T)\mathbf{y}
  =(\mathbf{x}^TL)\mathbf{y}
  =\mathbf{x}^T(L\mathbf{y})
  ={\langle}\mathbf{x},L\mathbf{y}{\rangle}
\end{equation*}
which shows that
\begin{equation*}
  {\langle}L\mathbf{x},\mathbf{y}{\rangle}
  ={\langle}\mathbf{x},L\mathbf{y}{\rangle},
\end{equation*}
exactly the self-adjoint condition satisfied by the S-L operator. 
\end{document}
