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A real symmetric matrix admitting a finite sum of nonnegative rank-one matrices $x_j{x}_j^T$. It is both a positive semidefinite matrix and a nonnegative matrix, but these two properties alone need not imply membership in the completely positive cone in arbitrary dimension. This matrix notion is distinct from a completely positive map.

A real symmetric matrix $A$ is copositive if its quadratic form satisfies $x^{T}Ax\ge 0$ for every $x$ in the nonnegative orthant. Every positive semidefinite matrix and every symmetric nonnegative matrix is copositive, as is their sum. The converse fails for the Horn copositive matrix.

The five-dimensional Horn copositive matrix has diagonal entries $1$, entries $-1$ on the edges of the five-cycle, and entries $1$ on the remaining pairs:For $x\ge 0$, a cyclic relabelling puts a smallest coordinate at $x_5$. The identitythen proves that $H$ is a copositive matrix.

However, $H$ is outside the positive-semidefinite-plus-nonnegative cone. Set $w=(1,2,1,0,0{)}^T$. Its quadratic form is zero. If $H=P+N$ with $P$ a positive semidefinite matrix and $N$ a symmetric nonnegative matrix, both $w^{T}Pw$ and $w^{T}Nw$ must vanish. Positivity of the first three coordinates of $w$ forces every entry of the leading $3\times 3$ block of $N$ to vanish. Applying the argument to all cyclic shifts of $w$ forces every entry of $N$ to vanish, since each pair of indices lies in a cyclic interval of length three. This would make $H$ a positive semidefinite matrix, but zero quadratic form of a positive semidefinite matrix would then give $Hw=0$, whereas $Hw=(0,0,0,2,2{)}^T$.

A square nonnegative matrix $A$ is irreducible if for each pair $(i,j)$ there is an integer $k\ge 0$ with $(A^k{)}_{ij}>0$. This says that every index can reach every other through positive entries. The Perron–Frobenius theorem then gives a positive leading eigenvector and an algebraically simple eigenvalue. Irreducibility alone permits other eigenvalues of the same modulus: the cyclic permutation matrix $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ has eigenvalues $1,-1$.

Let $S_{+}^{n-1}$ be the nonnegative portion of the unit sphere specified in the paper. If $Ax=0$ for some $x\in S_{+}^{n-1}$, then this nonzero vector is an eigenvector with eigenvalue zero, which already proves the assertion.

Otherwise $Ax\ne 0$ for every $x\in S_{+}^{n-1}$, andis a continuous map from $S_{+}^{n-1}$ to itself. Indeed, every entry of $Ax$ is nonnegative because $A$ is a nonnegative matrix and every entry of $x$ is nonnegative. By the supplied homeomorphism $h:{\mathbb{B}}^{n-1}\to S_{+}^{n-1}$, the continuous map $h^{-1}Th$ maps the finite-dimensional closed unit ball into itself. The Brouwer fixed-point theorem supplies a fixed point, hence $T(x_{\ast })=x_{\ast }$ for some $x_{\ast }\in S_{+}^{n-1}$. ThereforeTogether with the zero case this proves that $A$ has a nonnegative eigenvalue, with a nonnegative eigenvector. For $n=1$, the auxiliary closed unit ball is the one-point space ${\mathbb{B}}^0$ and the argument remains valid. No claim of strict positivity of every entry of the eigenvector is needed.

The necessary and sufficient condition iswhere $\rho$ is the spectral radius of the nonnegative matrix. No irreducibility is required. We use the given nonnegative-vector characterization of its Perron eigenvalue.

For necessity, suppose $p\ne 0$ is the survival fixed point and let $S=\{i:p_i>0\}$. Since $1-e^{-z}<z$ for every $z>0$, its equation gives $(\Lambda p{)}_i>p_i$ for every $i\in S$. There are finitely many such coordinates, soOutside $S$ the inequality $\Lambda p\ge \alpha p$ still holds, because its right side is zero and its left side is nonnegative. The provided Perron–Frobenius theorem characterization therefore gives $\rho (\Lambda )\ge \alpha >1$. In particular survival is impossible at $\rho =1$ for this Poisson offspring law.

Conversely suppose $\rho >1$. The given characterization supplies a nonzero $v\ge 0$ with $\Lambda v\ge \rho v$; normalize it so ${\max }_i{v}_i=1$. For sufficiently small $\eta >0$,For example any $0<\eta \le \min (1,2(\rho -1)/{\rho }^2)$ works. Coordinates with $v_i=0$ satisfy the inequality automatically. Starting at $x_0=\eta v$, iterate $x_{t+1}=T(x_t)$. Monotonicity makes the sequence increase, and $T$ takes nonnegative vectors into $[0,1{]}^l$, so it converges to a fixed point $x_{\infty }\ge \eta v\ne 0$. Part (a) gives $p\ge x_{\infty }$, proving survival from at least one type. This proves the spectral survival criterion for multitype Poisson branching for reducible as well as irreducible reproduction matrices.

For $x$ in the nonnegative orthant, the positive semidefinite matrix $P$ satisfies $x^{T}Px\ge 0$. The nonnegative matrix $N$ satisfiessince every summand is nonnegative. Adding these inequalities provesConsequently $A=P+N$ is a copositive matrix. The symmetry of $N$ follows from that of $A$ and $P$; entrywise nonnegativity alone need not imply positive semidefiniteness.

This gives the inclusion of the positive-semidefinite-plus-nonnegative cone in the copositive cone. It is only a sufficient construction; the argument does not claim that every copositive matrix has this decomposition in arbitrary dimension.

First suppose $A=P+N$, with $P$ a real positive semidefinite matrix and $N$ a symmetric nonnegative matrix. Factor $P=B^{T}B$, and write $v(z)=(z_1^2,\dots ,z_n^2{)}^T$. ThenEach term is a square multiplied by a nonnegative scalar, so $p$ is a sum of squares polynomial.

Conversely, suppose $p={\sum }_{\ell }{q}_{\ell }^2$. Because $p$ is a degree-four homogeneous polynomial, the homogeneous sum of squares representation allows every $q_{\ell }$ to be quadratic and homogeneous. Explicitly, higher-degree parts cannot cancel in a sum of squares; constant parts vanish because $p(0)=0$, and the degree-two part ${\sum }_{\ell }(q_{\ell }^{(1)}{)}^2$ forces all linear parts to vanish. WriteSince $p$ is invariant under every coordinate sign change, sign averaging of a sum of squares over independent Rademacher random variables givesThe cross terms vanish because their sign products contain an odd power of at least one independent sign. SetThen $P$ is a positive semidefinite matrix and $N$ is a symmetric nonnegative matrix. Comparing the coefficients of $z_i^4$ and $z_i^2{z}_j^2$ gives $A=P+N$. This proves the sum of squares criterion for a biquadratic form.

Let $w=(1,2,1,0,0{)}^T$. Direct multiplication gives $Hw=(0,0,0,2,2{)}^T$, so $w^{T}Hw=0$.

Suppose $H=P+N$, where $P$ is a positive semidefinite matrix and $N$ is a symmetric nonnegative matrix. Since $w\ge 0$, both $w^{T}Pw$ and $w^{T}Nw$ are nonnegative. Their sum is zero, so both vanish. In particular,Every coefficient $w_i{w}_j$ in this sum is strictly positive and every $N_{ij}$ is nonnegative, forcing $N_{ij}=0$ for $1\le i,j\le 3$.

Now apply the same argument to all five cyclic shifts of $w$. The Horn copositive matrix is cyclically invariant, so every shifted vector also has zero quadratic form. It follows that the entries of $N$ vanish on every cyclic block of three consecutive indices. Every pair of indices on a five-cycle lies in such a block, hence $N=0$.

This would imply $P=H$. But the zero quadratic form of a positive semidefinite matrix would then force $Hw=0$, contradicting $Hw=(0,0,0,2,2{)}^T$. Therefore $H$ lies outside the positive-semidefinite-plus-nonnegative cone. By the previous equivalence, its quartic form is a nonnegative polynomial that is not a sum of squares polynomial.

The null space consists of concentration patterns on which transport produces no change. Here it is the familyEvery box has the same concentration. Ocean circulation may continue, but inflow and outflow of the chemical tracer balance in each box. With $\mathbf{b}=0$ these are exactly the steady states; physically admissible members have $\alpha \ge 0$. With nonzero $\mathbf{b}$, adding $\alpha 1$ to any particular steady state leaves its transport balance unchanged, but a null space vector alone need not balance the sources and sinks.

To identify the tracer inventory, apply conserved inventory from a left null vector. Directly summing each column givesConsequently, for the supplied ordinary differential equations,For $\mathbf{b}=0$, $M$ is a conserved quantity. The connected box model equalizes concentrations, approaching the unique member of the null space with the initial inventory:The arithmetic mean is appropriate here because the supplied matrix has zero column sums. In a general box model the conserved quantity is $w^T\mathbf{c}$, where $w$ is a positive left null space vector; for physical concentrations expressed per unit volume, these weights represent the box volumes up to a common factor. One must establish the weights from the equations, rather than infer equal volumes from a schematic drawing.

Nonnegative initial concentrations remain nonnegative under transport alone. Indeed $A+2I$ is a nonnegative matrix, sohas nonnegative entries for $t\ge 0$: it is a positive matrix semigroup. Its column sums are one by $1^{T}A=0$. To see why equalization occurs, set $d=\mathbf{c}-(M/4)1$ for $\mathbf{b}=0$; then $1^{T}d=0$ andThe right side vanishes only for constant $d$, which in the zero-sum vector subspace means $d=0$. Its strict negative bound on the unit sphere in that vector subspace gives decay to zero. Thus the matrix exponential redistributes inventory without creating negative concentrations; arbitrary prescribed sinks can still exhaust the tracer if they are allowed to act indefinitely.

The sums $P+N$ of a real positive semidefinite matrix $P$ and a symmetric nonnegative matrix $N$ form a convex cone inside the copositive cone. Both terms have nonnegative quadratic forms on the nonnegative orthant. The Horn copositive matrix shows that the inclusion is strict in dimension five; the sum of squares criterion for a biquadratic form explains this cone's relation to semidefinite programming.

A square nonnegative matrix $A$ is primitive if some positive integer power $A^k$ has strictly positive entries. It is therefore an irreducible nonnegative matrix. The Perron–Frobenius theorem gives a leading eigenvalue whose modulus is strictly larger than that of every other eigenvalue. Every strictly positive matrix is primitive; the two-cycle permutation matrix is irreducible but not primitive.

This criterion holds even for reducible nonnegative matrices. If a fixed point $p\ne 0$ exists, $1-e^{-z}<z$ for $z>0$ gives $(\Lambda p{)}_i>p_i$ on its support. Some $\alpha >1$ therefore satisfies $\Lambda p\ge \alpha p$, forcing the spectral radius above one by the Perron–Frobenius theorem. Conversely take $v\ge 0$, $v\ne 0$, with $\Lambda v=\rho v$ and $\rho >1$. For sufficiently small $\eta >0$, $1-e^{-\eta \rho v_i}\ge \eta v_i$. Thus $\eta v$ is a subsolution; monotone iteration under $T$ gives a positive fixed point below $1$. Strictness of $1-e^{-z}<z$ excludes survival at the critical value for this Poisson offspring law.