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%TCIDATA{<META NAME="Title" CONTENT="Reference/Mathematics/Diagonalization">}
%TCIDATA{Created=Mon Aug 19 14:52:24 1996}
%TCIDATA{LastRevised=Wed Feb 12 15:48:44 1997}
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\section{Diagonalization\label{Diagonalization}}

\subsubsection{\protect\vspace{1pt}Diagonal Matrix\label{diagonal matrix}}

Let $A$ be an $n\times n$ matrix that has $n$ linearly independent
eigenvectors. Let $S$ be the matrix whose columns are the eigenvectors of $A$%
. Then $S^{-1}AS$ is a diagonal matrix $\Lambda $ with the eigenvalues of $A$
along its diagonal: 
\[
S^{-1}AS=\Lambda =\left[ 
\begin{array}{llll}
\lambda _{1} &  &  &  \\ 
& \lambda _{2} &  &  \\ 
&  & \ddots &  \\ 
&  &  & \lambda _{n}
\end{array}
\right] 
\]

\begin{itemize}
\item  Any matrix with distinct eigenvalues can be diagonalized since the $n$
eigenvectors corresponding to these eigenvalues will be independent.

\item  The diagonalizing matrix $S$ is not unique. Any column of $S$ can be
multiplied by a nonzero constant to produce a new diagonalizing matrix.

\item  The equation $SA=\Lambda S$ only holds if the columns of $S$ are the
eigenvectors of $A$. Other matrices $S$ will not produce a diagonal matrix $%
\Lambda $.

\item  Not all matrices are diagonalizable since not all matrices possess $n$
linearly independent eigenvectors.

\item  If the eigenvectors $x_{1},...,x_{k}$ correspond to different
eigenvalues $\lambda _{1},...,\lambda _{k}$, then those eigenvectors are
linearly independent.

\item  Let $A$ be an $n\times n$ matrix with eigenvectors $\lambda
_{1},...,\lambda _{n}$. Then the eigenvalues of $A^{k}$ are $\lambda
_{1}^{k},...,\lambda _{n}^{k}$, the $k^{th}$ powers of the eigenvalues of $A$%
. \ Each eigenvector of $A$ is also an eigenvector of $A^{k}$. If $S$
diagonalizes $A$, it also diagonalizes $A^{k}$: 
\[
\Lambda ^{k}=\left( S^{-1}AS\right) \left( S^{-1}AS\right) \cdots \left(
S^{-1}AS\right) =S^{-1}A^{k}S
\]

\item  If $A$ and $B$ are diagonalizable, they share the same eigenvector
matrix $S$ if and only if $AB=BA$.
\end{itemize}

\subsubsection{Multiplicity\label{multiplicity}}

\begin{definition}
The multiplicity of $\lambda $ as a root of the \hyperref{characteristic
polynomial}{}{}{MA05_01.tex#characteristic equation} is called the \textbf{%
algebraic multiplicity} of $\lambda $. The dimension of the \hyperref{%
eigenspace}{}{}{MA05_01.tex#eigenvectors} $E\left( \lambda \right) $ is
called the \textbf{geometric multiplicity} of $\lambda $.
\end{definition}

\begin{theorem}
Let $A$ be an $n\times n$ matrix and let $\lambda _{1},...,\lambda _{r}$ be
the distinct eigenvalues of $A$. Then $A$ is diagonalizable if and only if
the geometric multiplicity of each eigenvalue is equal to its algebraic
multiplicity.
\end{theorem}

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