\documentclass{article}
%
\usepackage[mathcal,mathbf]{euler}
\usepackage{theorem,amsmath,enumerate,fancyhdr,amssymb,accents,amsfonts}
\usepackage[pdftex]{graphicx}
\usepackage{myDefs}
\usepackage{algorithm2e}
\usepackage{enumitem}
\newtheorem{df}{Define}
\newtheorem{ex}{Example}
%In order to be consistent use the following notation in your notes:
%
% Here is where the title goes. Write down lecture number for ?
\title{
Semidefinite and Second Order Cone Programming Seminar\\
Fall 2012\\
Lecture 10 %1 or 2 or 3 etc
}
\author{
{\bf Instructor:} Farid Alizadeh\\
{\bf Scribe:} Joonhee Lee
}
\date{11/19/2012} %put the date of the lecture here NOT the date the note was written
\begin{document}
\pagestyle{fancy}
\lhead{{\bf scribe: }{Joonhee Lee}
\\{\bf Lecture 10} } %insert lecture number
\rhead{{\bf Date: }11/19/2012} %enter lecture date NOT today's date
\maketitle
%
%to be covered in the lecture
%\section{}
%Use sections and subsections as needed
%\oddsidemargin 0.0in
%\textwidth 6.25in
%\topmargin -0.25in
%\textheight 8.25in
%\begin{document}
%\begin{center}
%\large{ \bf Semidefinite and 2-order Cone Programming}\\
%\large{Fall 2001,lecture 2}\\
%\large{Instructor: Farid Alizadeh}\\
%\large{Scribe: Xuan Li} \\
%\end{center}
\medskip
\section{Overview} %give a one paragraph short description of the topics
In this lecture, we show that a number of sets and functions related to
the notion of \emph{sum-of-squares (SOS)} are SD-representable.
We will start with positive polynomials. Then, we introduce a general
algebraic framework in which the notion of sum-of-squares can be
formulated in very general setting.
\section{Polynomials}
Recall the cone of nonnegative univariate polynomials:
\[
\Po_{2d}[t]=\bigl\{p(t) = p_0 + p_1t + p_2t^2 +\cdots +
p_{2d}t^{2d}\geqslant 0 \quad \forall{t} \in \mathbb{R}
\]
Earlier we have examined this cone and have shown that it is
SD-representable. We now consider the case of multivariate polynomials.
The set of nonnegative polynomials is
\begin{displaymath}
\Po_{n,d} [t_1,\ldots,t_d] = \bigl\{\
P(t_1,\ldots,t_n) \geq 0 \qquad \forall{t} \in \mathbb{R} \bigr\}
\end{displaymath}
Recall that in the case of univariate polynomials, a polynomial
$p(t)\geqslant 0$ for all $t\in\real$ if and only if there are two
polynomials $p_1(t), p_2(t)$ such that $p(t)=p_1^2(t) + p_2^2(t)$. In
other words, a univariate polynomial is nonnegative over the real line
if and only if it is a sum of squares. For multivariate polynomials this
is no longer true.
\begin{Example}{(Motzkin Polynomial)}
consider the following polynomial:
\[
p(x,y) = \mathit{x}^4\mathit{y}^2 + \mathit{x}^2\mathit{y}^4 - 3\mathit{x}^2\mathit{y}^2 +1
\]
This polynomial is nonnegative for all $x,y\in \real$ because
\[
\frac{x^4y^2 + x^2y^4 + 1}{3} \geq \sqrt[3]{\mathit{x}^6\mathit{y}^6}
\]
by the arithmetic-geometric inequality. On the other hand it cannot be a
sum of two square polynomials. If there were polynomials $p_1(x,y),
\ldots, p_n(x,y)$ such that $\sum_ip_i^2(x,y)= p(x,y)$, then we must
have that each $p_i(x,y)$ is of the form
\[
x^2\bigl(a_iy^2+b_iy+c_i\bigr) + x(d_iy^2+e_iy+f_i) + g_iy^2+h_iy+k_i
\]
But it is immediate that $\sum a_i^2=\sum c_i^2=\sum f_i^2=\sum
g_i^2=\sum h_i^2=0$, and thus, $a_i=c_i=f_i=g_i=h_i=0$ for all $i=1,\ldots,n$,
otherwise the sum of squares
will contains terms which do not appear in $p(x,y)$. This leaves us with
\[
p(x,y)=\sum_i\bigl(b_ix^2y+d_ixy^2+e_ixy+k_i\bigr)^2
\]
But this implies that $\sum_ie_i^2=-3$ which is impossible.
This completes the proof that
the Motzkin polynomial is not sum of squares of other
polynomials. \hfill \rule{1mm}{1mm}
\end{Example}
Define
\[
\Sigma_{n,d}[t_1,\ldots,t_d]\defeq\bigl\{ p(t_1,\ldots,t_n)\mid
p(t_1,\ldots,t_n)=\sum_i^N p_i^2(t_1,\ldots,t_d)\}
%P(t_1,\ldots,t_n)\quad \mid
%\begin{array}{ll}
% &P\textrm{ is a polynomial and } \exists\,\, P_1,P_2,\ldots,P_n\\
%& \textrm{such that } P=P_1^2+\cdots+P_n^2 \\
%\end{array} \right\}
\]
\medskip
It is clear that $\Sigma_{n,2d} \subseteq \Po_{n,2d}$, since any
sum-of-square polynomial is nonnegative. The Motzkin example shows that
the inclusion is proper.
%However, the converse is not true by the following counter example.\\
When can we guarantee $P$$ $= $\Sigma$? It turns out that the following
are the only possible cases where $\Po=\Sigma$:
\begin{enumerate}
\item \textbf{Case 1:$n=1$.} This is the case of univariate polynomials which we have
proved earlier
\item \textbf{Case 2: $d$=2.} All polynomials of degree two can be written in the form
of $\vt^\top A\vt +\vb^\top\vt + c$, with $A\succ 0$ ($A$ is
nonsingular). To see this, first
suppose $\vd$ is a vector satisfying $\vd=B^{-T}\vb$, and let $A=B^\top B$ Then
\[
\vt^\top B^\top B\vt +\vb^\top\vt + c =
\bigl(B\vt+\tfrac{\vd^\top\vt}{2}\bigr)^\top\bigl(B\vt+\tfrac{\vd^\top\vt}{2}\bigr)
\]
which is a sum-of-squares.
\item\textbf{Case 3: $d=4,n=3$.} A very special case is three-variable
polynomials of degree four, where nonnegative polynomials
are always sum-of-squares. The proof is somewhat hairy.
\end{enumerate}
%For univariate polynomial : degree 2 polynomial ($\mathit{x}^TQ\mathit{x}$ $\Leftrightarrow$ $\mathit{x}^TP^TP\mathit{x}$)\\
%$\indent$For multivariate polynomial : polynomials of degree 4 with 3 variables.
Hilbert's seventeenth problem states that all nonnegative polynomials
are sum-of-squares of \emph{rational functions}. In other words for each
$p(\vt)=p(t_1,\ldots,t_n)\geqslant 0$ for all $t_i\geqslant 0$, there are
polynomials $p_1(\vt), \ldots, p_N(\vt), q(\vt)$ such that
\[
q^2(\vt)p(\vt) = p_1^2(\vt) + \cdots + p_N^2(\vt)
\]
%\begin{displaymath}
%\begin{array}{ll}
%\textrm{Consider } P(t_1,\ldots,t_n) \geq 0 \quad \\
%\Leftrightarrow \exists \textrm{ polynomials }q, q_1, \ldots, q_k \textrm{ such that}
%q^2p = q_1^2 + \cdots + q_k^2 \Leftrightarrow p = (\frac{q_1}{q})^2+\cdots+(\frac{q_k}{q})^2
%\end{array}
%\end{displaymath}
Hilbert's seventeenth problem was proved by E. Artin in 1923, and in the
process laid out the foundation of a field of algebra known as
\emph{real algebraic geometry}.
Assuming that the greatest common denominator of $q(\vt)$ and
$p_i(\vt)$ is 1, there is still no satisfactory bound on the number $N$,
the number of squares, and $D$, the largest degree among $q(\vt)$ and
$p_i(\vt)$. In fact, bounds that are known for $N$ and $D$ are not only
depend on $n$, the number of variable, and $d$ the degree of $p(\vt)$,
but also the coefficients of $p$ as well.
%There are infinitely many possibilities to choose deg(q) and k (No bound exists). However, bound of deg(q) depends on the minimum degree of p.
\section{General Algebra}
Consider (A, B, $\diamond$) where $\diamond$ is a bilinear operator that
is $\diamond$ : A $\times$ A $\to$ B.
A and B are finite-dimensional real linear spaces
with $\dim{A} = m$ and $\dim(B) = n$.
Note that bilinearity assumption is equivalent to the distributive law:
\begin{itemize}
\item $a\diamond ( \alpha b+ \beta c) = \alpha a \diamond b + \beta a \diamond c$
\item $(\alpha b+ \beta c) \diamond a = \alpha b \diamond a + \beta c \diamond a$
\end{itemize}
Note also that bilinearity means that there are matrices $Q_i$ such that
$(a\diamond b)_i$ = $a^TQ_ib$. Indeed, to each element $\va\in A$ we can
associate a linear transformation $L_\va$ mapping $A \to B$, that is
$L_\va \vb = \va\diamond\vb$. The linear transformation $L_\va$ may be
represented by a $n\times m$ matrix (also written as $L\va$, whose
entries are linear forms in $a_i$.
Our object of interest is the following set:
\[
\Sigma_\diamond = \bigl\{\sum{\va_i\diamond \va_i}\mid \, \va_i \in
A\bigr\} \subseteq B.
\]
which we call the \emph{sum-of-squares cone} (or the SOS cone)
associated with the algebra.
Note that $\Sigma_\diamond$ is a convex cone, since adding to sums of
squares creates another sum of squares.
\begin{Example}
Suppose $A=\Po_d[t]$, the set of degree $d$ univariate polynomials, and
$B=\Po_{2d}[t]$, the set of degree $2d$ univariate polynomials. Then
the $(\Po_d[t], \Po_{2d}[t], *)$ forms an algebra, with $*$ indicating the
multiplication of polynomials. If we represent each polynomial by the
vector of its coefficients, then $*$ is the \emph{convolution}
operation:
\[
(p_0,p_1,\ldots, p_d)*(q_0,q_1,\ldots,q_d) = (p_0q_0, p_0q1+p_1q_0,
\ldots, p_0q_k+p_1q_{k-1} +\cdots, \ldots, p_dq_d)
\]
For this algebra $\Sigma_*$ is the set of polynomials of degree $2d$
which are sum of squares of polynomials. As we know, in this case this
cone is exactly the cone of polynomials of degree $2d$ which are
nonnegative for every $t\in \real$.
\end{Example}
We make three assumptions on the algebra $(A,B,\diamond)$ without loss
of generality, which will make the presentation cleaner and more
streamlined.
\begin{enumerate}
\item \textbf{$\diamond$ is commutative:} $\va \diamond \vb = \vb
\diamond \va$.
If $\diamond$ it is not commutative, then we can replace it with its
\emph{anti-commutator}:
\[
\va \overline{\diamond} \vb = \frac{\va \diamond \vb + \vb \diamond \va}{2}
\]
Note that for the algebras $(A, B, \diamond)$ and $(A, B,
\overline\diamond)$ we have $\Sigma_\diamond =
\Sigma_{\overline\diamond}$.
\item $B = \Span(A \diamond A)$
This assumption ensures that $B$ does not contain elements that are not
somehow generated by elements from $A$, and in turn results in
$\Sigma_\diamond$ to be full dimensional in $B$.
\item \textbf{The mapping $ L : A \to \mathbb{R}^{n\times m}$
is injective.}
\[L_{\vx_1} = \,L_{\vx_2} \Rightarrow \vx_1 =\vx_2.\]
Without the this assumption, there is no way to distinguish $\vx_1$ and
$\vx_2$. In this case we observe that the relation $L_{\vx_1}=L_{\vx_2}$
defines an equivalence relation on $A$:
\[\vx_1 \simeq \vx_2 \,\Leftrightarrow \,L_{\vx_1} = L_{\vx_2}\]
Then, by replacing $A$, with $A/\simeq$, the set of equivalence classes,
and defining on $A/\simeq$:
\[ [\vx]\diamond \vy] \defeq [\vx\diamond \vy] \]
Using commutativity it is easy to see that this definition is
consistent, that is, if $\vx_1\simeq\vx_2$ and $\vy_1\simeq\vy_2$ then
$\vx_1\diamond\vy_1 \simeq \vx_2\diamond\vy_2$. Therefore,
$(A/\simeq,B,\diamond)$ satisfies the third assumption.
\end{enumerate}
\begin{lemma}
With assumptions 1, 2, and 3,
\begin{enumerate}
\item $\Sigma_\diamond$ is full dimensional convex cone.
\item Every element in $\Sigma_\diamond$ is sum of at most $n =\dim(B)$
squares.
\end{enumerate}
\end{lemma}
\begin{proof}\\
1) Since the sum of two sums of squares is a another sum of square,
$\Sigma_\diamond$ is convex. To prove it is full-dimensional we claim
that $B = \Sigma_\diamond - \Sigma_\diamond$. First note that for any
$\va,\vb\in A, \va\diamond\vb=(\tfrac{a+b}{2})^{\diamond 2} -
(\frac{a-b}{2})^{\diamond 2}$; this follows from commutativity. On the
other hand by assumption 2, every element in $B$ is of the form $\sum_i
\va_i\diamond\vb_i$. This shows that $B=\Sigma_\diamond -
\Sigma_\diamond$, and thus, $\Sigma$ is full dimensional.
2) By Caratheodory's Theorem for cones, every element of
$\Sigma_\diamond$ is sum of at most $n$ extreme rays. But the extreme
rays of $\Sigma$ are among perfect squares $\va\diamond\va$, so each
element of $\Sigma_\diamond$ is sum of at most $n$ squares.
\end{proof}
\phantom{ }
\begin{flushleft}
(A Trivial Example) A = B = $\mathbb{C}$ : Complex number under ordinary multiplication. Then, $\Sigma_\diamond = \mathbb{C}$\\
\end{flushleft}
\begin{lemma}
If (A, B, $\diamond$) is formally real$\footnote{$\sum_{i}a_i^{\diamond
2} = 0 \Rightarrow a_i$ = 0}$,then $\Sigma_\diamond$ is pointed. Conversely, if $\Sigma_\diamond$ is pointed and there are no nilpotent$\footnote{$a_i^{\diamond 2}$ = 0}$ elements, then (A, B, $\diamond$) is formally real.
\end{lemma}
\begin{proof}
\begin{eqnarray*}
&&If\, -\sum_{i}a_i^{\diamond 2} \in \Sigma_\diamond, then\\
&\quad& \exists\, b_i : \sum_{i}b_i^{\diamond 2} = -\sum_{i}a_i^{\diamond 2} \Rightarrow \sum_{i}b_i^{\diamond 2} + \sum_{i}a_i^{\diamond 2} = 0\\
&\Rightarrow& a_i = 0 \,\,and\,\, b_i = 0 \,\,\textrm{(by formally real)}\\
&\quad&\\
&\quad& \sum_{i}a_i^{\diamond 2}=0 \Rightarrow a_i^{\diamond 2}=\vzero \Rightarrow a_i = 0 \,\,\textrm{(by nilpotent)}
\end{eqnarray*}
\end{proof}
The dual of $\Sigma_\diamond$ is $\Sigma_\diamond^\ast = \{\,z \mid \langle a , z \rangle \ge 0, \forall a \in\,\Sigma_\diamond \}$. Then,
\begin{theorem}
$\Sigma_\diamond$ is a proper cone iff\, $\Sigma_\diamond^\ast$ is.
\end{theorem}
\begin{df}
$\Lambda$ and $\Lambda^\ast$ operators
\begin{eqnarray*}
&\quad& (a,b,\diamond), \Sigma_\diamond, \Sigma_\diamond^\ast\\
&& \Lambda_\diamond : B \to \mathbb{S}_A \text{ where }\mathbb{S}_A\,\, \text{is a set of symmetric bilinear forms}\\
&& (\Lambda_\diamond(w),a,b) \triangleq \langle w, a\diamond b \rangle_B\\
&&a^T\Lambda b = b^T\Lambda a, (a,b) \in A, w \in B\\
\end{eqnarray*}
\end{df}
\begin{theorem}
$w \in \Sigma_\diamond^\ast$ iff $\Lambda_\diamond(w) \succeq$ 0.
\end{theorem}
\begin{proof}
$\Rightarrow:$
\begin{eqnarray*}
&& \Lambda_\diamond (w) \succeq 0\\
&\quad& \Lambda_\diamond (w)(a, a) = \langle w, a\diamond a \rangle \ge 0\\
&\quad& (\because w \in \Sigma_\diamond^\ast\,\, and\,\, a\diamond a \in \Sigma_\diamond )\\
&\quad&\ \Rightarrow \Lambda (w) \ge 0\\
\end{eqnarray*}
$\textrm{Next, we need to show}$ ($\Leftarrow$)
\begin{eqnarray*}
\Lambda (w) \ge 0 &\Rightarrow& \forall a \in A \,\, \text{such that} \,\, \Lambda_\diamond (w)(a, a) \ge 0\\
&\Rightarrow& \langle w, x \rangle \ge 0 \forall x \in \Sigma_\diamond\\
&\Rightarrow& w \ \in\, \Sigma_\diamond^\ast
\end{eqnarray*}
\end{proof}
Note that, if $\Lambda^\ast$ : $\mathbb{S} \to B$, then
\[ \langle \mathit{x}, \Lambda (w) \rangle_{\mathbb{S}_A}\,\, =\,\, \langle \mathit{x} , \Lambda^\ast (x) \rangle_B \quad \,\,\forall w \in B\,\, and\,\, \mathit{x} \in \mathbb{S}_A \]
\begin{theorem}
$u \in \Sigma_\diamond\,$ iff $\exists \mathit{Y} \succeq 0$ such that
$u = \Lambda^\ast (\mathit{Y})$.
\end{theorem}
\begin{proof}\\
Show ($\Rightarrow$)
\begin{eqnarray*}
&\quad& if\,\, \mathit{Y} \succeq 0 \,\, and\,\, \Lambda^\ast (mathit{Y} ) = u, \,\, then\,\, \\
&& \forall v \in \Sigma_\diamond^\ast,\,\, \langle u, v \rangle_B = \langle \Lambda^\ast (\mathit{Y}, v \rangle_B = \langle \mathit{Y}, \Lambda{v} \rangle_{\mathbb{S}_A} \ge 0\\
&& (\because \mathit{Y} \ge 0 \,\, and \,\, \Lambda(v) \ge 0)\\
&\Rightarrow& u \in \Sigma_\diamond^{\ast \ast} = \Sigma_\diamond\\
\end{eqnarray*}
($\Leftarrow$)
\[ \text{if } u \in \Sigma_\diamond, \,\, then\,\, \exists a_i \text{ such that } u = \sum_{i} a_i^{\diamond 2}.\,\, Let\,\, v \in \mathit{B} \\ \]
\begin{eqnarray*}
\langle u, v \rangle_B &=& \langle \sum_{i} a_i^{\diamond 2}, v \rangle_B = \sum_{i} \langle a_i^{\diamond 2}, v \rangle_B\\
&=& \sum_{i} \Lambda (v)(a_i, a_i) = \sum_{i} \langle \Lambda (v), a_ia_i^T \rangle_{\mathbb{S}_A}, \\
&\overset{\text{ (Let } \mathit{Y} = \sum_{i} a_ia_i^T)}{=}& \langle \Lambda (v), \mathit{Y} \rangle_{\mathbb{S}_A} = \langle v, \Lambda^\ast (\mathit{Y} \rangle_B\\
&\Rightarrow& u = \Lambda^\ast (\mathit{Y})
\end{eqnarray*}
\end{proof}
\begin{Example}
%Suppose A $\in \mathbb{R}^n$ and $(\mathbb{R}^{n+1}, \mathbb{R}^{2n+1}, *)$
%\begin{equation*}
%\begin{pmatrix}
%p_0 \\
%p_1 \\
%\vdots \\
%p_n
%\end{pmatrix}
%*
%\begin{pmatrix}
%q_0 \\
%q_1 \\
%\vdots \\
%q_n
%\end{pmatrix}
%=
%\begin{pmatrix}
%p_0q_0 \\
%p_0q_1 +p_1q_0\\
%p_0q_2 + p_1q_1 + p_2q_0\\
%\vdots \\
%p_nq_n
%\end{pmatrix}\\
%\end{equation*}
For the algebra of univariate polynomials, the $\Lambda$ operator can be
computed as follows. If we represent $\Lambda(\vw)$ as a matrix, then
its $i,j$ entry, by definition, is given by $\langle \vw,
\ve_i*\ve_j\rangle$. But, $\ve_i*\ve_j = \ve_{i+j}$, thus
$\bigl(\Lambda(\vw)\bigr)_{ij}=w_{i+j}$. This means that the $i,j$ entry
of $\Lambda$ in this case only depends on $i+j$, that is, all entries of
$\Lambda(\vw)$ with the same $i+j$ must be equal. It follows that
$\Lambda$ has all entries on the reverse diagonals are equal.
\[
\Lambda(\vw)= \begin{pmatrix}
w_0 & w_1 & \cdots & w_n\\
w_1 & w_2 & \cdots & w_{n+1}\\
\vdots & \vdots & \ddots & \vdots \\
w_n & w_{n+1} & \cdots & w_{2n}\\
\end{pmatrix}
\]
\end{Example}
\subsection{Squared functional systems}
Of particular interest are algebras that are induced by
\emph{functional spaces}. Let
$\mathcal{F}$ = $\{f_1, f_2, \cdots, f_m\}$ where each $f_i :
\Delta \to \mathbb{R}$, is a real-valued function.
Let $F = \Span(\mathcal{F}) = \sum_{i}\alpha_i f_i(x), \quad \forall x
\in \Delta$ be the linear space spanned by $f_i$, where $(\alpha_if_i +
\alpha_jf_j)(x) = \alpha_if_i(x) + \alpha_jf_j(x)$.
Now define
$\mathcal{S} = \{f_i, f_j\}$ and $(f_i, f_j)(x) = f_i(x) f_j(x)$\\
Let $S = \Span(\mathcal{I}$)
$(F, S, \dot)$ is an algebra and $\Sigma_\mathcal{F} = \{ \sum_{i}g_i^2
: g_i \in F\}$. The algebra $(F,S,\cdot)$ along with its SOS cone
$\Sigma_\mathcal{F}$ is called \emph{a squared function of system.}
The univariate polynomial example given earlier is a special case where
$\F=\{1,t,t^2,\ldots, t^d\}$, and $S=\{1,t,t^2,\ldots,t^{2d}\}$.
\subsection{The semidefinite and second order cones as SOS cones}
For two $p\times q$ matrices $A$ and $B$ define the \emph{Cracovian
multiplication} as follows:
\[
A \diamond B = AB^T \text{ and } A\overline{\diamond} B =
\tfrac{A\diamond B+ B\diamond A}{2}
\]
Then, for the algebra $(\real^{p\times q}, \real^{p\times p},
\overline{\diamond})$, the SOS cone $\Sigma_\diamond$ is exactly the
cone of positive semidefinite $p\times p$ matrices.
%\begin{eqnarray*}
%&\textcircled{1}& (\mathbb{R}^{mxn},\, S_m,\, \bar{\diamond}) \Rightarrow \Sigma_\diamond = \{\sum_{i}A_iA_i^T\} = \mathcal{P}\\
%&\textcircled{2}& (\mathbb{R}^{m},\,S_m,\, \bar{\diamond}) \Rightarrow \Sigma_\diamond = \{\sum_{i}v_iv_i^T\}\\
%&\textcircled{3}& (S_m,\,S_m,\,\circ) \,\,where\,\, A\circ B = \frac{AB+BA}{2} \Rightarrow \Sigma_\circ = \mathcal{P}_{nxn}\\
%\end{eqnarray*}
\section{Operations on algebras and their SOS cones}
\subsection{Bijective linear transformations}
Let $(A,B,\diamond)$ be an algebra which as usual satisfies assumptions
1,2, and 3, and let $C$ be another linear space,
such that the linear transformation $F\mid B\to C$ is bijective. (Note
that this means that necessarily $\dim(B)=n=\dim(C)$). Define a new
binary operation $\circ\mid A\times A\to C$ by $L_\circ = FL_\diamond$.
Then $(A,C, \circ)$ is an algebra, and if satisfying assumptions 1,2, and 3.
Furthermore, $\Sigma_\circ=F \Sigma_\diamond$.
\begin{definition}
Two cones $K_1$ and $K_2$ are linearly isomorphic if $\exists$ bijective linear transformation F : $ K_1 = \mathit{F}K_2$.
\end{definition}
Thus, if $\Sigma_1$ is an SOS cone, and $\Sigma_2$ is a cone linearly
isomorphic to $\Sigma_1$ then $\Sigma_2$ is also an SOS cone for some
algebra.
%$\Sigma_\diamond \leftrightarrow$ (A, B, $\diamond$) and F $\:$ B $\to$ B is bijective. Then,\\
%$\phantom{ad}$ F$\Sigma_\diamond$ is also an sum-of-square (SOS) cone.\\
%$\leftrightarrow$ (A, B, $\diamond_1$), a$\,\diamond_1$\,b $\triangleq$ $L_{\diamond_1}$(a)b = F$L_\diamond$(a)b
\subsection{Isomorphism and linear isomorphism among algebras}
Let $(A_1, B_1, \diamond)$ and $(A_2, B_2, \diamond)$ be two algebras,
and let there be two linear transformation $F$ and $G$ such that
\begin{displaymath}
\left. \begin{array}{ll}
F : A_1 \to A_2 & \\
G : B_1 \to B_2 &
\end{array} \right\}
\end{displaymath}
If both $F$ and $G$ are bijective, and we have
\[
G(a\,\diamond_1 b) = F(a)\, \diamond_2 F(b)
\]
we say that these algebras are \emph{isomorphic}. Note that as opposed
to ordinary algebraic structures we need two maps two define
isomorphism.
\begin{lemma}
If $(A_1,B_1, \diamond_1)$ and $(A_2,B_2, \diamond_2)$ are two algebras
isomorphic to each other, then their SOS cones $\Sigma_{\diamond_1}$ and
$\Sigma_{\diamond_2}$ are linearly isomorphic.
\end{lemma}
Let $y\in\Sigma_{\diamond_2}$. Then
\begin{gather*}
y=\sum_iy_i\diamond_2
y_i=\sum_iF(x_i)\diamond_2F(x_i)\quad\text{(for some $x_i\in A$, since $F$ is
surjective)}\\
=\sum_i G(x_i\diamond_1 x_i)\quad \text{(by definition of
homomorphism)}\\
=G\Bigl(\sum_ix_i\diamond_1 x_i\Bigr) \in
G\bigl(\Sigma_{\diamond_1}\bigr) \quad \text{(by linearity)}.
\end{gather*}
The sequence of implications above goes through in both directions, establishing
that $\Sigma_{\diamond_2}=G\bigl(\Sigma_{\diamond_1}\bigr)$. By definition,
if $G$ is bijective, then it is also a linear isomorphism between
$\Sigma_{\diamond_1}$ and $\Sigma_{\diamond_2}$.
\subsection{Direct sums of algebras}
For $k$ algebras $(A_1,B_1,\diamond_1), \cdots, (A_k,B_k,\diamond_k)$
define a new algebra
\[
(A_1 \times \cdots \times A_k, B_1 \times \cdots \times B_k, \diamond)
\]
with
\begin{displaymath}
\begin{pmatrix}
a_1 \\
\vdots\\
a_k
\end{pmatrix}
\diamond
\begin{pmatrix}
b_1 \\
\vdots\\
b_k
\end{pmatrix}
=
\begin{pmatrix}
a_1 \diamond_1 b_1\\
\vdots\\
a_k \diamond_k b_k
\end{pmatrix}
\end{displaymath}
This new algebra is called the \emph{direct sum algebra}. It is
immediate that
\begin{displaymath}
\Sigma_\diamond = \Sigma_{\diamond_1} \times\cdots\times
\Sigma_{\diamond_k}\\
\end{displaymath}
And the $\Lambda$ operator is given by
\begin{displaymath}
\Lambda_\diamond\bigl((\vw_1,\ldots,\vw_k)\bigr) =
\Lambda_1\oplus\cdots\oplus\Lambda_k.
\end{displaymath}
\subsection{Minkowski sum of algebras}
Consider the algebras $(A_1, B, \diamond_1), \cdots, (A_k, B,
\diamond_2)$ with possible different $A_i$, but all having the same $B$. The
\emph{Minkowski sum algebra} is the algebra $(A_1\times\cdots\times
A_k,B,\diamond)$, with
\begin{displaymath}
\begin{pmatrix}
a_1 \\
\vdots\\
a_k
\end{pmatrix}
\diamond
\begin{pmatrix}
b_1 \\
\vdots\\
b_k
\end{pmatrix}
= a_1 \diamond_1 b_1 + \cdots +a_k\diamond_k b_k
\end{displaymath}
Then we have
\[
\Sigma_\diamond = \Sigma_{\diamond_1} + \cdots +\Sigma_{\diamond_k}
\text{ and }
\Lambda_\diamond(\vw) = \Lambda_{\diamond_1}(\vw) + \cdots + \Lambda_{\diamond_k}(\vw)
\]
\subsection{Weighted sum-of squares}
Combining the Minkowski sum and linear transformations we can show that
a kind of \emph{Weighted Sums Of Squares (WSOS)} is also in fact a
sum-of squares. This follows from the following observation: Let $(A_i,
B_i, \diamond_i)$, for $\i=1,\ldots, k$ be algebras, and let $F_i$ be linear
transformations each mapping $B_i$ to a common set $B$. Then the cone
\[
F_1\Sigma_{\diamond_1} + \cdots + F_k\Sigma_{\diamond_k}
\]
is also an SOS cone. Here is an example:
%\begin{ex}
%\begin{displaymath}
%\Lambda_\diamond = \Lambda_{\diamond_1} + \Lambda_{\diamond_2}
%where \,\, \Lambda = \Lambda_1 \oplus \Lambda_2 =
%\begin{pmatrix}
%\Lambda_1 & 0\\
%0 & \Lambda_2
%\end{pmatrix}
%\end{displaymath}
%\end{ex}
\begin{Example}
Let
$\mathbb{P}_{[0,\infty)}^{\,d}$ = $\{\,p_d(t)\,\mid\,p(t) \ge \forall t
\ge 0 \,\}$, where $\deg(\mathbb{P}) = d$. Clearly this set is a convex
cone. Let us show that it is in fact ans SOS cone.
\begin{displaymath}
\mathcal{P}_{[0,\infty)}^{\,d} =
\left\{ \begin{array}{ll}
\mathcal{P}^d + t\mathcal{P}^{d-2}, &\text{d is even}\\
\mathcal{P}^{d-1} + t\mathcal{P}^{d-1}, &\text{d is odd}\\
\end{array} \right.
\end{displaymath}\\
\begin{proof}\\
We have : $p(t) \ge 0 \,\, \forall t \ge\,0 \,\Leftrightarrow \,p(t^2)
\ge 0\,\,\forall t \in \mathbb{R}$. Thus, $q(t) = p(t^2) = p_1^2(t) +
p_2^2(t)$ for some polynomials $p_1$ and $p_2$. We have,
%$\phantom{By using facts, we know : }\,\,\,p(t^2) = p_1^2(t) + p_2^2(t)$. Then,
%\begin{displaymath}
\begin{align*}
p(t^2) &= q^2(t)+r^2(t)\quad\text{separating odd and even degree
terms in $q(t)$ and $r(t)$:}\\
&=\bigl(q_1(t^2) + t q_2(t^2)\bigr)^2 +
\bigl(r_1(t^2)+tr_2(t^2)\bigr)^2\\
&=q_1^2(t^2)+t^2q_2^2(t^2)+\underbrace{2tq_1(t^2)q_2(t^2)}_{=0 } +
r_1^2(t^2)+t^2r_2(t^2) + \underbrace{2tr_1(t^2)r_2(t^2)}_{=0}\\
&\indent (\divideontimes\,\,
2tq_1(t^2)q_2(t^2)=2tr_1(t^2)r_2(t^2)=0\text{ since all terms of
$p(t^2)$ have even degree})\\
&\Rightarrow p(t) = p_1^2(t) + tp_2^2(t) \text{ (By changing $t^2$ to $t$)}
\end{align*}
%\end{displaymath}
\end{proof}
Thus, we have shown that the cone $\Po_{[0,\infty]}^d[t]$ is a weighted
sum of squares. However, note that the operation of multiplying by $t$
is a bijective linear transformation mapping the space of degree $d$
polynomials, to the space of degree $d+1$ polynomials with a zero
constant term. Thus, $t\Po_\real[t]$ is an SOS cone, and, its
Minkowski sum $\Po[t] + t\Po[t]$ is also an SOS
cone\footnote{More precisely, if $d$ is even then
$\Po_{[0,\infty]}^d[t]=\Po^d[t] +t\Po^{d-2}[t]$, and if $d$ is odd
$\Po_{[0,\infty]}^d[t] = \Po^{d-2}[t] + t\Po^{d-1}[t]$.}.
\end{Example}
\subsection{Isomorphism by change of basis and change of variables}
Our presentation of algebras has been \emph{basis-free} in that all
arguments are made independent of any particular basis for $A$ or $B$ in
the algebra $(A,B,\diamond)$. Of course in practice, the multiplication
operator $L_\diamond$ is represented by a matrix, and this
representation in turn depends on the particular basis chosen for $A$
and $B$. When we change basis for $A$, it is tantamount to replacing
$L_\diamond(\vx)$ with $L_\diamond(F\vx)$ where $F$ is the change of
basis matrix. Similarly changing basis in $B$ is the same as replacing
$L_\diamond(\vx)$ with $L_\diamond(\vx)G$, where $G$ is the change of
basis matrix in $B$. Needless to say, the resulting algebras are all
isomorphic to each other, and thus, the resulting SOS cones are linearly
isomorphic.
Polynomials, and in general squared functional systems, are functional
linear spaces. As such, there are many different ways of choosing a basis
for them. For instance, for polynomials, in the ordinary representation
$p_0 + p_1t + \cdots + p_dt^d$ we are using the basis
$\{1,t,t^2,\ldots,t^d\}$. However, there are many other bases: for
instance $\{1, t+1, (t+1)^2,\ldots, (t+1)^d\}$ is another basis, and use
of orthogonal polynomials (such as Chebychev, Legendre, Laguerre, etc.)
are other ways of representing polynomials. Clearly, changing the basis
in which polynomials are represented does not affect the SOS cone, nor
does it change the fact that it equals the cone of nonnegative polynomials
in the univariate case.
The second observation is the effect of change of variable even in a
nonlinear way.
In general, consider the set of polynomials of degree $d$ which are
nonnegative over a set $\Delta\subseteq\real$. Let $H:\Omega\to\Delta$
be an onto mapping from a set $\Omega$ to
$\Delta$. Note that $\Omega$ need not be a subset of $\real$; it is
entirely arbitrary. Then the cone
\[
\Po_\Omega[H]=\{f(x)\mid f(x)= p_0 + p_1H(x) + \ldots + p_dH^d(x) \geq 0
\forall x\in \Omega\}
\]
is a convex cone linearly isomorphic to $\Po_\Delta[t]$.
\begin{eqnarray*}
f(x) \ge 0\,\, &\Leftrightarrow&\, p_0 + p_1H(x) + \cdots + p_{d}H^{d}(x) \ge 0\quad \forall x \in \Omega\\
&\Leftrightarrow&\, p(t) \ge 0\quad \forall x \in \Delta\\
\end{eqnarray*}
This means that
from polynomials, we can construct other sets of SOS cones by functional
composition and possibly by change of basis. Two examples follow:
\begin{Example} Consider the set of polynomials which are
nonnegative over a finite interval say $[0,1]$.
$\mathcal{P}_{[0,1]}^{\,d}$ = $\{\,p_d(t)\,\mid\,p(t) \ge \forall t \in [0, 1] \,\}$\\
%
%Consider, $\mathcal{P}^{\,2d}[t]$\\
%
%Let H : $\Omega \to \Delta$ and\\
%$\indent \{ f(x) : \Delta \to \mathbb{R} : f(x) = p_0 + p_1H(x) + \cdots + p_{2d}H^{2d}(x)\}$\\
%$\indent$ Then, $\mathcal{P}_\Delta^{\,2d}$ and $\mathcal{P}_\Omega^{\,H,2d}$ are linearly isomorphic ($\mathcal{P}_\Delta^{\,2d}\,\, \cong\,\, \mathcal{P}_\Omega^{\,H,2d}$)\\
%\\
%(Why?)
%\end{Example}
%
%\begin{ex}
Setting $H(t) = \frac{t}{1+t}$, we see that $H: [0,\infty]\to[0,1]$.
Then a polynomial $p(t) \geq 0$ for all $t\in[0,\infty]$ iff
$p(s)\geq0$ for all $s\in[0,\infty]$, and this is equivalent to
$p\bigl(\tfrac{t}{1+t}\bigr)\geq 0 $ for all $t\in[0,1]$. But expanding
this function, and observing that multiplying it by $(1+t)^d$ does not
change its sign over $[0,\infty]$, we see that
\begin{displaymath}
\forall t \in [\,0,\infty)
\,\,p\left(\frac{t}{1+t}\right)= \frac{q(t)}{(1+t)^d} \ge 0,
\Leftrightarrow \quad \forall t \in [0,1] \Leftrightarrow q(t)\ge 0\quad\forall t\in[0,1]
\end{displaymath}
This implies that $\Po_{[0,1]}[t] \simeq \Po_{[0,\infty]}[t]$, and both
are weighted SOS, and thus SOS cones.
\end{Example}
\begin{Example} \emph{Cosine polynomials} are functions of the form
$p_0+p_1\cos(t) + \cdots + p_d\cos(dt)$. We are interested in the cone
\[
\Po_{\cos}[t]=\{\vp \mid p(t)=p_0+p_1\cos(t) + \cdots p_d\cos(td)\ge 0
\forall t\in\real\}
\]
Since the domain of the function $\cos$ is the set [-1,1], and since
each $\cos(kt)$ is a polynomial of degree $k$ in $\cos(t)$ (known as the
Chebychev polynomials of the first kind), we conclude that
\[
\Po_{\cos}[t]\simeq \Po_{[-1,1]}[t] \simeq \Po_{[0,1]}[t] \simeq
\Po_{[0,\infty]}[t]
\]
\end{Example}
\begin{thebibliography}{9}
\bibitem{nes1} Nesterov, Y. \textit{Squared functional systems and
optimization problems}, In \textit{High Performance Optimization}, pp.
405-440, Kluwer Acad. Publ., Dordrecht, 2000.
\bibitem{fayb1} Faybusovich, L. \textit{On te Nesterov's Approach to
Semi-Infinte Programming}, Acta. Appli. Math., 74, (2002), pp. 195-215.
\bibitem {Alizadeh11} Papp, D, and Alizadeh, F., \textit{Semidefinite
Characterization of Sum-Of-Squares Cones in Algebras}, RUTCOR-rrr Report
no 11-14, RUTCOR, Rutgers-State University of New Jersey, accepted for
publication in \emph{SIAM J. on Optimization}.
\end{thebibliography}
\end{document}