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\begin{document}


\section{\label{Smith normal form}Special Forms of Matrices}

Several normal forms of matrices can be obtained in \textsl{Scientific
Notebook}.

\subsection{\ Smith Normal Form}

Two $n\times n$ matrices $A$ and $B$ are \textsl{equivalent }if 
\index{Equivalent matrices@Equivalent matrices} 
\index{Matrices@Matrices!equivalent@equivalent}one can be obtained from the
other by a sequence of elementary row and column operations. In other words, 
$B=QAP$ for some invertible matrices $Q$ and $P$.

Any matrix $A$ with polynomial entries is equivalent to a diagonal matrix of
the form $diag(1,\ldots ,1,p_{1},p_{2},\ldots ,p_{k},0,\ldots ,0)$, where $%
0\leq \deg (p_{1})\leq \deg (p_{2})\leq \ldots \leq \deg (p_{k})$ and for
each $i$, $p_{i}$ is a factor of $p_{i+1}$. This matrix, which is uniquely
determined by $A$, is called the \textsl{Smith normal form} of $A$. Two
square matrices with polynomial entries are equivalent if and only if they
have the same 
\index{Matrices@Matrices!Smith normal form@Smith normal form}\label%
{Matrices, Smith normal form}Smith normal form. 
\index{Smith normal form@Smith normal form}

\begin{quote}
$\blacktriangleright $ \textsf{Matrices + Smith Normal Form}
\end{quote}

\begin{quotation}
$A=\left[ 
\begin{array}{cc}
x^{3}+3x-1 & x^{2}-3x+5 \\ 
x^{3}+5x^{2} & x^{4}+3x^{2}
\end{array}
\right] $ 
\[
\text{Smith normal form: }\left[ 
\begin{array}{cc}
1 & 0 \\ 
0 & x^{7}+5x^{5}-3x^{4}+19x^{3}-28x^{2}
\end{array}
\right] 
\]

\ 

$B=\left[ 
\begin{array}{cc}
97 & 50 \\ 
79 & 56
\end{array}
\right] \left[ 
\begin{array}{cc}
x^{3}+3x-1 & x^{2}-3x+5 \\ 
x^{3}+5x^{2} & x^{4}+3x^{2}
\end{array}
\right] \left[ 
\begin{array}{cc}
-85 & -55 \\ 
-37 & -35
\end{array}
\right] $, 
\[
\text{Smith normal form: }\left[ 
\begin{array}{cc}
1 & 0 \\ 
0 & x^{7}+5x^{5}-3x^{4}+19x^{3}-28x^{2}
\end{array}
\right] 
\]
\end{quotation}

\ 

Two $n\times n$ matrices $A$ and $B$ are \textsl{similar } 
\index{Similar matrices@Similar matrices}\label{Similar matrices}if there 
\index{Matrices@Matrices!similar@similar}\label{Matrices, similar} is an
invertible $n\times n$ matrix $C$ such that $B=C^{-1}AC$.\medskip

\begin{example}
Two matrices are similar if and only if their characteristic matrices $xI-A$
and $xI-B$ are equivalent, which is the same as saying that $xI-A$ and $xI-B$
have the same Smith normal form. To demonstrate this relationship, take two
similar matrices.

$A=\left[ 
\begin{array}{cc}
1 & 2 \\ 
3 & 4
\end{array}
\right] $

$B=\left[ 
\begin{array}{rr}
1 & 9 \\ 
-3 & 4
\end{array}
\right] ^{-1}\left[ 
\begin{array}{cc}
1 & 2 \\ 
3 & 4
\end{array}
\right] \left[ 
\begin{array}{rr}
1 & 9 \\ 
-3 & 4
\end{array}
\right] =\,\left[ 
\begin{array}{rr}
\dfrac{61}{31} & \vspace{6pt}-\dfrac{319}{31} \\ 
-\dfrac{24}{31} & \,\dfrac{94}{31}
\end{array}
\right] $ \vspace{6pt}

These matrices have the following characteristic matrices

$xI-A=\left[ 
\begin{array}{cc}
x & 0 \\ 
0 & x
\end{array}
\right] -\left[ 
\begin{array}{cc}
1 & 2 \\ 
3 & 4
\end{array}
\right] =\,\left[ 
\begin{array}{cc}
x-1 & -2 \\ 
-3 & x-4
\end{array}
\right] \vspace{6pt}$

$xI-B=\left[ 
\begin{array}{cc}
x & 0 \\ 
0 & x
\end{array}
\right] -\left[ 
\begin{array}{rr}
\dfrac{61}{31} & -\dfrac{319}{31}\vspace{6pt} \\ 
-\dfrac{24}{31} & \dfrac{94}{31}\,
\end{array}
\right] =\,\left[ 
\begin{array}{cc}
x-\dfrac{61}{31}\vspace{6pt} & \dfrac{319}{31} \\ 
\dfrac{24}{31} & x-\dfrac{94}{31}
\end{array}
\right] $ \vspace{6pt}

with Smith normal forms both equal to\vspace{6pt} 
\[
\left[ 
\begin{array}{cc}
1 & 0 \\ 
0 & x^{2}-5x-2
\end{array}
\right] 
\]
\bigskip
\end{example}

\subsection{\ \label{Companion matrix}Rational Canonical Form}

\label{Polynomials, companion matrix}The \textsl{companion matrix} of 
\index{Companion matrix@Companion matrix}a monic 
\index{Matrices@Matrices!companion@companion}\label{Matrices, companion}%
polynomial $a_{0}+a_{1}X+\cdots +a_{n-1}X^{n-1}+X^{n}$ of degree $n$ is the $%
n\times n$ matrix with a subdiagonal of ones, final column 
\[
\left[ 
\begin{array}{cccc}
-a_{0} & -a_{1} & \cdots & -a_{n-1}
\end{array}
\right] ^{T} 
\]
and other entries zero.

\begin{quote}
$\blacktriangleright $ \textsf{Polynomials + Companion Matrix}
\end{quote}

\begin{quotation}
$x^{4}+3x^{2}-2x+1$, Companion matrix: $\left[ 
\begin{array}{cccc}
0 & 0 & 0 & -1 \\ 
1 & 0 & 0 & 2 \\ 
0 & 1 & 0 & -3 \\ 
0 & 0 & 1 & 0
\end{array}
\right] \medskip $

$x^{3}+ax^{2}+bx+c$, Companion matrix: $\left[ 
\begin{array}{ccc}
0 & 0 & -c \\ 
1 & 0 & -b \\ 
0 & 1 & -a
\end{array}
\right] $
\end{quotation}

Note that the first of the following matrices is the companion matrix of its
own characteristic and minimum polynomials.

\ 

\begin{quote}
$\blacktriangleright $ \textsf{Matrices + Minimum Polynomial}
\end{quote}

\begin{quotation}
$\left( \ 
\begin{array}{ccccc}
0 & 0 & 0 & 0 & -a \\ 
1 & 0 & 0 & 0 & -b \\ 
0 & 1 & 0 & 0 & -c \\ 
0 & 0 & 1 & 0 & -d \\ 
0 & 0 & 0 & 1 & -e
\end{array}
\right) $, minimum polynomial: 
\[
\quad a+bX+cX^{2}+dX^{3}+eX^{4}+X^{5} 
\]
\end{quotation}

Choosing \textsf{Rational Canonical Form} from\ the \textsf{Matrices}
submenu produces a factorization of a square matrix as $PFP^{-1}$, where $F$
is in rational canonical form. 
\index{Frobenius form@Frobenius form}\label{Frobenius form}A \textsl{%
rational } 
\index{Matrices@Matrices!rational canonical form@rational canonical form}%
\label{Matrices, rational canonical form}\textsl{canonical} \textsl{form}, 
\index{Rational canonical form@Rational canonical form}\label%
{Rational canonical form}sometimes called a\textsl{\ Frobenius form}, is a
block diagonal matrix with each block the companion matrix of its own
minimum and characteristic polynomials. Each of the minimum polynomials of
these blocks is a factor of the characteristic polynomial of the original
matrix. The polynomials that determine the blocks of the rational canonical
form sequentially divide one another.

\begin{quote}
$\blacktriangleright $ \textsf{Matrices + Rational Canonical Form}
\end{quote}

\begin{quotation}
$\left[ 
\begin{array}{ccc}
1 & 2 & 3 \\ 
4 & 5 & 6 \\ 
7 & 8 & 9
\end{array}
\right] =\,\left[ 
\begin{array}{rrr}
1 & 1 & 30 \\ 
0 & 4 & 66 \\ 
0 & 7 & 102
\end{array}
\right] \left[ 
\begin{array}{rrr}
0 & 0 & 0 \\ 
1 & 0 & 18 \\ 
0 & 1 & 15
\end{array}
\right] \left[ 
\begin{array}{rrr}
1 & -2\; & 
\vspace{6pt}1\; \\ 
0 & -\dfrac{17}{9} & \vspace{6pt}\dfrac{11}{9} \\ 
0 & \dfrac{7}{54} & -\dfrac{2}{27}
\end{array}
\right] $
\end{quotation}

Notice that the rational canonical form in the preceding example is the
companion matrix of its minimum polynomial $X^3-15X^2-18X$. Now look at the
companion matrix of this same matrix.

\begin{quote}
$\blacktriangleright $ \textsf{Evaluate}
\end{quote}

\begin{quotation}
$x\left( 
\begin{array}{ccc}
1 & 0 & 0 \\ 
0 & 1 & 0 \\ 
0 & 0 & 1
\end{array}
\right) -\left[ 
\begin{array}{ccc}
1 & 2 & 3 \\ 
4 & 5 & 6 \\ 
7 & 8 & 9
\end{array}
\right] =\left[ 
\begin{array}{ccc}
x-1 & -2 & -3 \\ 
-4 & x-5 & -6 \\ 
-7 & -8 & x-9
\end{array}
\right] \,\ $
\end{quotation}

\begin{quote}
$\blacktriangleright $ \textsf{Matrices + Smith Normal Form}
\end{quote}

\begin{quotation}
$\left[ 
\begin{array}{ccc}
x-1 & -2 & -3 \\ 
-4 & x-5 & -6 \\ 
-7 & -8 & x-9
\end{array}
\right] $, Smith normal form: $\left[ 
\begin{array}{ccc}
1 & 0 & 0 \\ 
0 & 1 & 0 \\ 
0 & 0 & -18x-15x^{2}+x^{3}
\end{array}
\right] $
\end{quotation}

The polynomial occurring in the preceding Smith normal form has its
coefficients displayed in the rational canonical form shown previously.

\ 

\begin{quote}
$\blacktriangleright $ \textsf{Matrices +} \textsf{Rational Canonical Form}
\end{quote}

\begin{quotation}
$\left[ 
\begin{array}{rrr}
5 & -6 & -6 \\ 
-1 & 4 & 2 \\ 
3 & -6 & -4
\end{array}
\right] =\,\left[ 
\begin{array}{rrr}
1 & 5 & 2 \\ 
0 & -1 & 0 \\ 
0 & 3 & 1
\end{array}
\right] \left[ 
\begin{array}{rrr}
0 & -2 & 0 \\ 
1 & 3 & 0 \\ 
0 & 0 & 2
\end{array}
\right] \left[ 
\begin{array}{rrr}
1 & -1 & -2 \\ 
0 & -1 & 0 \\ 
0 & 3 & 1
\end{array}
\right] $
\end{quotation}

There are two blocks in the preceding rational canonical form:

\begin{enumerate}
\item[1.]  The companion matrix $\left[ 
\begin{array}{cc}
0 & -2 \\ 
1 & 3
\end{array}
\right] $ of $X^{2}-3X+2=\,\left( X-1\right) \left( X-2\right) $

\item[2.]  The companion matrix $\left[ 2\right] $ of $X-2$
\end{enumerate}

\begin{quote}
$\blacktriangleright $ \textsf{Matrices + Characteristic Polynomial, Factor}
\end{quote}

\begin{quotation}
$\left[ 
\begin{array}{rrr}
5 & -6 & -6 \\ 
-1 & 4 & 2 \\ 
3 & -6 & -4
\end{array}
\right] $, characteristic polynomial: 
\[
X^{3}+8X-5X^{2}-4=\,\left( X-1\right) \left( X-2\right) ^{2} 
\]
\end{quotation}

\begin{quote}
$\blacktriangleright $ \textsf{Matrices + Minimum Polynomial, Factor}
\end{quote}

\begin{quotation}
$\left[ 
\begin{array}{rrr}
5 & -6 & -6 \\ 
-1 & 4 & 2 \\ 
3 & -6 & -4
\end{array}
\right] $, minimum polynomial: $2-3X+X^{2}=\,\left( X-1\right) \left(
X-2\right) $
\end{quotation}

The characteristic matrix $xI-A$ of the preceding matrix $A$ is 
\[
x\left[ 
\begin{array}{ccc}
1 & 0 & 0 \\ 
0 & 1 & 0 \\ 
0 & 0 & 1
\end{array}
\right] -\left[ 
\begin{array}{rrr}
5 & -6 & -6 \\ 
-1 & 4 & 2 \\ 
3 & -6 & -4
\end{array}
\right] =\,\left[ 
\begin{array}{ccc}
x-5 & 6 & \;6 \\ 
\;1 & x-4 & -2 \\ 
-3 & 6 & x+4
\end{array}
\right] 
\]

\begin{quote}
$\blacktriangleright $ \textsf{Matrices + Smith Normal Form}
\end{quote}

\begin{quotation}
$\,\left[ 
\begin{array}{ccc}
x-5 & 6 & \;\;6 \\ 
\;1 & x-4 & -2 \\ 
-3 & 6 & x+4
\end{array}
\right] $, Smith normal form: $\left[ 
\begin{array}{ccc}
1 & 0 & 0 \\ 
0 & x-2 & 0 \\ 
0 & 0 & 2-3x+x^{2}
\end{array}
\right] $
\end{quotation}

These two examples illustrate a relationship among the Smith normal form,
the characteristic matrix, and the rational canonical form of a
matrix.\medskip

\begin{quotation}
\emph{Note\quad }The Smith normal form of the characteristic matrix of $A$
displays the factors of the characteristic polynomial of $A$ that determine
the rational canonical form of $A.$
\end{quotation}

\ 

\subsection{\ Jordan Form\label{Jordan form}}

Choosing \textsf{Jordan Form } 
\index{Jordan form@Jordan form}from\ the \textsf{Matrices} 
\index{Matrices@Matrices!Jordan form@Jordan form}\label%
{Matrices, Jordan form}submenu produces a factorization of a square matrix
as $PJP^{-1}$, where $J$ is in Jordan form. This form is a block diagonal
matrix with each block an elementary Jordan matrix. More specifically, the 
\textsl{Jordan form} of an $n\times n$ matrix $A$ is a matrix of the form 
\[
J(A)=\left[ 
\begin{array}{cccc}
J_{n_{1}}\left( \lambda _{1}\right) & 0 & \cdots & 0 \\ 
0 & J_{n_{2}}\left( \lambda _{2}\right) & \cdots & 0 \\ 
\vdots & \vdots & \ddots & \vdots \\ 
0 & 0 & \cdots & J_{n_{k}}\left( \lambda _{k}\right)
\end{array}
\right] 
\]
where $n_{1}+n_{2}+\cdots +n_{k}=n$, and each diagonal block $%
J_{n_{i}}\left( \lambda _{i}\right) $ is an $n_{i}\times n_{i}$ \textsl{%
elementary Jordan matrix} of the form 
\index{Elementary Jordan matrix@Elementary Jordan matrix}\label%
{Elementary Jordan form} 
\[
J_{n_{i}}\left( \lambda _{i}\right) =\left[ 
\begin{array}{ccccc}
\lambda _{i} & 1 & \cdots & 0 & 0 \\ 
0 & \lambda _{i} & \cdots & 0 & 0 \\ 
\vdots & \vdots & \ddots & \vdots & \vdots \\ 
0 & 0 & \cdots & \lambda _{i} & 1 \\ 
0 & 0 & \cdots & 0 & \lambda _{i}
\end{array}
\right] 
\]

\begin{quote}
$\blacktriangleright $ \textsf{Matrices + Jordan Form}
\end{quote}

\begin{quotation}
$\left[ 
\begin{array}{rrr}
2 & -1 & 0 \\ 
-1 & 2 & -1 \\ 
0 & -1 & 2
\end{array}
\right] =$

\[
\left[ 
\begin{array}{ccc}
\frac{1}{2} & \vspace{5pt}\frac{1}{4} & \frac{1}{4} \\ 
0 & \vspace{5pt}\frac{1}{4}\sqrt{2} & -\frac{1}{4}\sqrt{2} \\ 
-\frac{1}{2} & \frac{1}{4} & \frac{1}{4}
\end{array}
\right] \left[ 
\begin{array}{ccc}
2 & 0 & 0 \\ 
0 & 2-\sqrt{2} & 0 \\ 
0 & 0 & 2+\sqrt{2}
\end{array}
\right] \left[ 
\begin{array}{ccc}
1 & 0 & -1 \\ 
1 & \sqrt{2} & 1 \\ 
1 & -\sqrt{2} & 1
\end{array}
\right] 
\]
\end{quotation}

Thus, the Jordan form is 
\[
J\left( \left[ 
\begin{array}{rrr}
2 & -1 & 0 \\ 
-1 & 2 & -1 \\ 
0 & -1 & 2
\end{array}
\right] \right) =\left[ 
\begin{array}{ccc}
2 & 0 & 0 \\ 
0 & 2-\sqrt{2} & 0 \\ 
0 & 0 & 2+\sqrt{2}
\end{array}
\right] 
\]
In this case, $J(A)$ is diagonal, so each $J_{n_i}\left( \lambda _i\right) $
is a $1\times 1$ matrix. The matrix 
\[
A=\left( 
\begin{array}{rrr}
2 & -1 & 0 \\ 
-1 & 2 & -1 \\ 
0 & -1 & 2
\end{array}
\right) 
\]
has characteristic and minimum polynomials 
\[
-4+10X-6X^2+X^3=\,\left( X-2\right) \left( X-2-\sqrt{2}\right) \left( X-2+%
\sqrt{2}\right) 
\]
whose roots $\left\{ 2,\;2+\sqrt{2},\;2-\sqrt{2}\right\} $ are the diagonal
entries of the Jordan form.

\begin{quote}
$\blacktriangleright $ \textsf{Matrices + Jordan Form}
\end{quote}

\begin{quotation}
$\left[ 
\begin{array}{cccc}
2 & 0 & 0 & 0 \\ 
1 & 2 & 0 & 0 \\ 
0 & 0 & 2 & 0 \\ 
0 & 0 & -3 & 2
\end{array}
\right] =\,\left[ 
\begin{array}{cccc}
0 & 1 & 0 & 0 \\ 
1 & 0 & 0 & 0 \\ 
0 & 0 & 0 & 1 \\ 
0 & 0 & -3 & 0
\end{array}
\right] \left[ 
\begin{array}{cccc}
2 & 1 & 0 & 0 \\ 
0 & 2 & 0 & 0 \\ 
0 & 0 & 2 & 1 \\ 
0 & 0 & 0 & 2
\end{array}
\right] \left[ 
\begin{array}{cccc}
0 & 1 & 0 & 0 \\ 
1 & 0 & 0 & 0 \\ 
0 & 0 & 0 & -\frac{1}{3} \\ 
0 & 0 & 1 & 0
\end{array}
\right] $
\end{quotation}

Thus, the Jordan form is$\,$%
\[
J\left( \left[ 
\begin{array}{cccc}
2 & 0 & 0 & 0 \\ 
1 & 2 & 0 & 0 \\ 
0 & 0 & 2 & 0 \\ 
0 & 0 & -3 & 2
\end{array}
\right] \right) =\left[ 
\begin{array}{cccc}
2 & 1 & 0 & 0 \\ 
0 & 2 & 0 & 0 \\ 
0 & 0 & 2 & 1 \\ 
0 & 0 & 0 & 2
\end{array}
\right] 
\]

In this case, $J_{n_1}\left( \lambda _1\right) =J_{n_2}\left( \lambda
_2\right) =\left[ 
\begin{array}{cc}
2 & 1 \\ 
0 & 2
\end{array}
\right] $, the companion matrix of the minimum polynomial of 
\[
A=\left[ 
\begin{array}{cccc}
2 & 0 & 0 & 0 \\ 
1 & 2 & 0 & 0 \\ 
0 & 0 & 2 & 0 \\ 
0 & 0 & -3 & 2
\end{array}
\right] 
\]
The matrix $A$ has characteristic polynomial $\left( X-2\right) ^4$ with
roots $\left\{ 2,\;2,\;2,\;2\right\} $, and minimum polynomial $%
X^2-4X+4=\,\left( X-2\right) ^2$.

\begin{quote}
$\blacktriangleright $ \textsf{Matrices + Jordan Form}
\end{quote}

\begin{quotation}
$\left[ 
\begin{array}{cccc}
2 & 0 & 0 & 0 \\ 
0 & 2 & 0 & 0 \\ 
0 & 0 & 2 & 0 \\ 
0 & 0 & 0 & 2
\end{array}
\right] \ =\,\left[ 
\begin{array}{cccc}
1 & 0 & 0 & 0 \\ 
0 & 1 & 0 & 0 \\ 
0 & 0 & 1 & 0 \\ 
0 & 0 & 0 & 1
\end{array}
\right] \left[ 
\begin{array}{cccc}
2 & 0 & 0 & 0 \\ 
0 & 2 & 0 & 0 \\ 
0 & 0 & 2 & 0 \\ 
0 & 0 & 0 & 2
\end{array}
\right] \left[ 
\begin{array}{cccc}
1 & 0 & 0 & 0 \\ 
0 & 1 & 0 & 0 \\ 
0 & 0 & 1 & 0 \\ 
0 & 0 & 0 & 1
\end{array}
\right] $
\end{quotation}

The preceding matrix is already in Jordan form. It has minimum polynomial $%
X-2$ and characteristic polynomial $\left( X-2\right) ^4$, the same
characteristic polynomial as the previous one, but a different minimum
polynomial and a different Jordan form.

\ 

\begin{quote}
$\blacktriangleright $ \textsf{Matrices + Jordan Form}
\end{quote}

\begin{quotation}
$\left( \left[ 
\begin{array}{rr}
1 & 2 \\ 
-1 & -1
\end{array}
\right] \right) =\left[ 
\begin{array}{cc}
\frac{1}{2}-\frac{1}{2}i & \vspace{6pt}\frac{1}{2}+\frac{1}{2}i \\ 
\frac{1}{2}i & -\frac{1}{2}i
\end{array}
\right] \left[ 
\begin{array}{cc}
i & 0 \\ 
0 & -i
\end{array}
\right] \left[ 
\begin{array}{cc}
1 & 1-i \\ 
1 & 1+i
\end{array}
\right] $
\end{quotation}

In this case, $J_{n_{1}}\left( \lambda _{1}\right) =\left[ i\right] $ and $%
J_{n_{2}}\left( \lambda _{2}\right) =\left[ -i\right] $ are $1\times 1$
matrices. The matrix $\left[ 
\begin{array}{rr}
1 & 2 \\ 
-1 & -1
\end{array}
\right] $ has characteristic and minimum polynomial $x^{2}+1=\left(
x+i\right) \left( x-i\right) .\medskip $

\subsection{\ \label{Orthogonal matrix}Orthogonal Matrices}

An \textsl{orthogonal matrix} is 
\index{Orthogonal matrix@Orthogonal matrix}a real 
\index{Unitary matrix@Unitary matrix}\label{Unitary matrix}matrix for which
the inner product of any two different columns is zero and the inner product
of any column with itself is one. The matrix is said to have 
\index{Orthonormal@Orthonormal}\label{Orthonormal columns} \textsl{%
orthonormal } 
\index{Matrices@Matrices!unitary@unitary}\label{Matrices, unitary}columns. 
\index{Matrices@Matrices!orthogonality test@orthogonality test}\label%
{Matrices, orthogonality test}Such a matrix necessarily has orthonormal rows
as well.

\begin{quote}
$\blacktriangleright $ \textsf{Matrices + Orthogonality Test}
\end{quote}

\begin{quotation}
$\left( 
\begin{array}{rr}
\cos \alpha & -\sin \alpha \\ 
\sin \alpha & \cos \alpha
\end{array}
\right) $, orthogonal? $true$

\ 

$\left( 
\begin{array}{crr}
0 & 0 & 1 \\ 
1 & 0 & 0 \\ 
0 & 1 & 0
\end{array}
\right) $, orthogonal? $true$

\ 

$\left( 
\begin{array}{cr}
0 & 1 \\ 
1 & 1
\end{array}
\right) $, orthogonal? $false\medskip $
\end{quotation}

\strut

\subsection{\ \label{SVD}Singular Value Decomposition (SVD)}

Any $m\times n$ real matrix $A$ can be factored into a product $A=UDV$, with 
$U$ and $V$ real orthogonal $m\times m$ and $n\times n$ matrices,
respectively, and $D$ a diagonal matrix with positive numbers in the first
rank-$A$ entries on the main diagonal and zeroes everywhere else. The
entries on the main diagonal of $D$ are 
\index{Singular values@Singular values}\label{Singular values}called the 
\index{Matrices@Matrices!singular values@singular values}\label%
{Matrices, singular values}\emph{singular values} of $A$. This factorization 
$A=UDV$ is called a \emph{singular value decomposition }of $A$. 
\index{Matrices@Matrices!singular value decomposition@singular value decomposition}%
\label{Matrices, singular value decomposition}

\begin{quote}
$\blacktriangleright $ \textsf{Matrices + Singular Values}
\end{quote}

\begin{quotation}
$\left[ 
\begin{array}{rrr}
5 & -5 & -3 \\ 
-3 & 0 & 5 \\ 
1.0 & 5 & 4
\end{array}
\right] $, singular values: $\left\{ 3.56,4.61,10.1\right\} $

\ 

$\left[ 
\begin{array}{ccc}
5 & -5 & -3 \\ 
-3 & 0 & 5
\end{array}
\right] $, singular values: $\left\{ 
\sqrt{14},\sqrt{79}\right\} =\left\{ 3.74,8.89\right\} $
\end{quotation}

\ 

\begin{quote}
\textsf{Matrices + SVD}
\end{quote}

\begin{quotation}
$\left[ 
\begin{array}{rrr}
5 & -5 & -3 \\ 
-3 & 0 & 5 \\ 
1 & 5 & 4
\end{array}
\right] =$%
\[
\left[ 
\begin{array}{rrr}
-.722 & -.191 & -.665 \\ 
.455 & .593 & -.664 \\ 
.522 & -.782 & -.341
\end{array}
\right] \left[ 
\begin{array}{ccc}
10.1 & 0 & 0 \\ 
0 & 4.61 & 0 \\ 
0 & 0 & 3.56
\end{array}
\right] \left[ 
\begin{array}{rrr}
-.443 & .618 & .649 \\ 
-.763 & -.640 & .0897 \\ 
-.471 & .456 & -.755
\end{array}
\right] 
\]
\end{quotation}

These two outer matrices fail the orthogonality test because they are
numerical approximations only. You can check the inner products of the
columns to see that they are ``approximately'' orthogonal.

\begin{quote}
$\blacktriangleright $ \textsf{Matrices + Singular Values, Matrices + SVD}
\end{quote}

\begin{quotation}
$\left( 
\begin{array}{cc}
1 & 2.0 \\ 
3 & 4
\end{array}
\right) $, singular values: $\left\{ .366,\;5.47\right\} $%
\[
\left( 
\begin{array}{cc}
1 & 2 \\ 
3 & 4
\end{array}
\right) =\left( 
\begin{array}{rr}
-.405 & -.915 \\ 
-.915 & .405
\end{array}
\right) \left( 
\begin{array}{cc}
5.47 & 0 \\ 
0 & .366
\end{array}
\right) \left( 
\begin{array}{rr}
-.576 & -.817 \\ 
.817 & -.576
\end{array}
\right) 
\]
\end{quotation}

\strut

\subsection{\ PLU Decomposition\label{Matrices, PLU decomposition}}

Any $m\times n$ real matrix $A$ can be factored into a product $A=PLU$, with 
$L$ and $U$ real lower and upper triangular $m\times m$ and $m\times n$
matrices, respectively, with $1$'s on the main diagonal of $L$, and with $P$
a permutation matrix. This factorization $A=PLU$ is called the \textsl{PLU
decomposition}\emph{\ }of $A$. 
\index{Matrices@Matrices!singular value decomposition@singular value decomposition}%
The matrix $U$ is an echelon form of $A$.

\begin{quote}
$\blacktriangleright $ \textsf{Matrices + PLU Decomposition}
\end{quote}

\begin{quotation}
$\left[ 
\begin{array}{rrr}
1 & 2 & 3 \\ 
2 & 4 & 6 \\ 
3 & 2 & 1
\end{array}
\right] =\allowbreak \left[ 
\begin{array}{ccc}
1 & 0 & 0 \\ 
0 & 0 & 1 \\ 
0 & 1 & 0
\end{array}
\right] \left[ 
\begin{array}{ccc}
1 & 0 & 0 \\ 
3 & 1 & 0 \\ 
2 & 0 & 1
\end{array}
\right] \left[ 
\begin{array}{ccc}
1 & 2 & 3 \\ 
0 & -4 & -8 \\ 
0 & 0 & 0
\end{array}
\right] \medskip $

$\left[ 
\begin{array}{cccc}
2 & -5 & -3 & -5 \\ 
7 & 5 & 9 & 6 \\ 
4 & 3 & 5 & -9
\end{array}
\right] \medskip =\allowbreak \left[ 
\begin{array}{ccc}
1 & 0 & 0 \\ 
0 & 1 & 0 \\ 
0 & 0 & 1
\end{array}
\right] \left[ 
\begin{array}{ccc}
1 & 0 & 0 \\ 
\frac{7}{2} & 1 & 0 \\ 
2 & \frac{26}{45} & 1
\end{array}
\right] \left[ 
\begin{array}{cccc}
2 & -5 & -3 & -5 \\ 
0 & \frac{45}{2} & \frac{39}{2} & \frac{47}{2} \\ 
0 & 0 & -\frac{4}{15} & -\frac{566}{45}
\end{array}
\right] $

$\left[ 
\begin{array}{cc}
45 & -8 \\ 
-93 & 92 \\ 
43 & -62
\end{array}
\right] =\allowbreak \left[ 
\begin{array}{ccc}
1 & 0 & 0 \\ 
0 & 1 & 0 \\ 
0 & 0 & 1
\end{array}
\right] \left[ 
\begin{array}{ccc}
1 & 0 & 0 \\ 
-\frac{31}{15} & 1 & 0 \\ 
\frac{43}{45} & -\frac{1223}{1698} & 1
\end{array}
\right] \left[ 
\begin{array}{cc}
45 & -8 \\ 
0 & \frac{1132}{15} \\ 
0 & 0
\end{array}
\right] \medskip $

$\left[ 
\begin{array}{cc}
.532 & 1.95 \\ 
1.5 & .0013
\end{array}
\right] =\allowbreak \left[ 
\begin{array}{cc}
0 & 1.0 \\ 
1.0 & 0
\end{array}
\right] \left[ 
\begin{array}{cc}
1.0 & 0 \\ 
.\,\allowbreak 354\,67 & 1.0
\end{array}
\right] \left[ 
\begin{array}{cc}
1.\,\allowbreak 5 & .00\,\allowbreak 13 \\ 
0 & 1.\,\allowbreak 949\,5
\end{array}
\right] \medskip $

$\left( 
\begin{array}{cc}
0 & 0 \\ 
0 & 0
\end{array}
\right) \medskip =\allowbreak \left( 
\begin{array}{cc}
1 & 0 \\ 
0 & 1
\end{array}
\right) \left( 
\begin{array}{cc}
1 & 0 \\ 
0 & 1
\end{array}
\right) \left( 
\begin{array}{cc}
0 & 0 \\ 
0 & 0
\end{array}
\right) $
\end{quotation}

Note that the upper triangular matrix in the first example is the same as
the following.

\begin{quote}
$\blacktriangleright $ \textsf{Matrices + Fraction-Free Gaussian Elimination}
\end{quote}

\begin{quotation}
$\left[ 
\begin{array}{rrr}
1 & 2 & 3 \\ 
2 & 4 & 6 \\ 
3 & 2 & 1
\end{array}
\right] $, fraction-free Gaussian elimination: $\left[ 
\begin{array}{ccc}
1 & 2 & 3 \\ 
0 & -4 & -8 \\ 
0 & 0 & 0
\end{array}
\right] $
\end{quotation}

\label{Gaussian elimination}In general, the upper triangular matrix in the
PLU decomposition is the echelon form of the original matrix obtained by
Gaussian elimination.\medskip

\subsection{\ Related topics}

\begin{itemize}
\item  \hyperref{Working With Matrices}{}{}{DM8-1.tex#Matrix Algebra}

\item  \hyperref{Standard Operations}{}{}{DM8-2.tex#Standard matrix
operations}

\item  \hyperref{Row Operations and Echelon Forms}{}{}{DM8-3.tex#Matrices,
row operations}

\item  \hyperref{Systems of Linear Equations}{}{}{DM8-4.tex#Systems of
equations}

\item  \hyperref{Matrix Equations}{}{}{DM8-4.tex#Matrix equations}

\item  \hyperref{Matrix Operators}{}{}{DM8-5.tex#Trace}

\item  \hyperref{Polynomials and Vectors Associated with a Matrix}{}{}{%
DM8-6.tex#Characteristic, polynomial}

\item  \hyperref{Vector Spaces Associated with a Matrix}{}{}{DM8-7.tex#Row
space}

\item  \hyperref{Exercises and Solutions}{}{}{DM8-9.tex#Exercises}
\end{itemize}

\rule{0.67in}{0.01in}

\hyperref{\FRAME{itbpF}{0.5275in}{0.3399in}{0.0519in}{}{}{index.wmf}{\special%
{language "Scientific Word";type "GRAPHIC";display "PICT";valid_file
"F";width 0.5275in;height 0.3399in;depth 0.0519in;original-width
36.125pt;original-height 22.5625pt;cropleft "0";croptop "1";cropright
"1";cropbottom "0";filename 'index.wmf';file-properties "XNPEU";}}}{}{}{%
DMIndex.tex#Math Index}\quad {\small Index entries: \hyperref{Matrices}{}{}{%
DMIndex.tex#Matrices} }

\end{document}
