A symmetric matrix can be positive in two different senses: positive semidefinite, and nonnegative entry by entry. Requiring both gives the cone of doubly nonnegative matrices
\[\mathcal D_n=\{X\in\mathbb R^{n\times n}:X=X^\top,\ X\succeq0,\ X_{ij}\geq0\}.\]A seemingly similar condition is to ask for a Gram factorization with nonnegative vectors. This defines the cone of completely positive matrices. The two cones agree up to dimension four — but not in dimension five.
The two cones
Write
\[\mathcal C_n= \{AA^\top:A\in\mathbb R_+^{n\times p} \text{ for some finite }p\}.\]Every such matrix is positive semidefinite and entrywise nonnegative, so \(\mathcal C_n\subseteq\mathcal D_n\). The converse holds for \(n\leq4\), and fails for every \(n\geq5\); see Anstreicher, Burer, and Dür’s study of the five-dimensional gap.
The terminology matters: “totally positive” usually refers to positivity of all minors, which is a different property from complete positivity.
A five-cycle counterexample
Set \(a=(\sqrt5-1)/2\) and consider
\[X= \begin{pmatrix} 1&a&0&0&a\\ a&1&a&0&0\\ 0&a&1&a&0\\ 0&0&a&1&a\\ a&0&0&a&1 \end{pmatrix}.\]Its nonzero off-diagonal entries trace the edges of a pentagon.
Theorem 1. The matrix \(X\) is doubly nonnegative but not completely positive: \(X\in\mathcal D_5\setminus\mathcal C_5\).
Proof: the five-cycle spectrum and a trace obstruction
Step 1: it is doubly nonnegative
Entrywise nonnegativity is immediate. To check the spectrum, write \(X=I+aH\), where \(H\) is the adjacency matrix of the five-cycle. Its Fourier eigenvectors give the eigenvalues
\[\lambda_k(X)=1+2a\cos\left(\frac{2\pi k}{5}\right), \qquad k=0,\ldots,4.\]The smallest cosine is \(-(1+\sqrt5)/4\). Our choice of \(a\) makes the smallest eigenvalue zero, so all eigenvalues are nonnegative.
Step 2: it is not completely positive
Suppose, for a contradiction, that
\[X=\sum_{\ell=1}^p u_\ell u_\ell^\top, \qquad u_\ell\geq0.\]Whenever \(X_{ij}=0\), nonnegativity forces \((u_\ell)_i(u_\ell)_j=0\) for every \(\ell\). Thus the support of each column \(u_\ell\) must be a clique of the five-cycle. A pentagon has no triangles, so each support contains at most two vertices, and those vertices must be adjacent.
For a vector supported on one edge, the elementary inequality \(s^2+t^2\geq2st\) yields
\[\|u_\ell\|^2 \geq 2\sum_{\{i,j\}\in E}(u_\ell)_i(u_\ell)_j,\]where \(E\) consists of the five unordered edges of the cycle. The same inequality holds for a vector supported on a single vertex.
Summing over the factor columns would imply
\[\operatorname{tr}X\geq 2\sum_{\{i,j\}\in E}X_{ij}=10a.\]But \(\operatorname{tr}X=5\) and \(a>1/2\). This contradiction proves \(X\notin\mathcal C_5\).
Corollary 1. For every \(n\geq5\), the inclusion \(\mathcal C_n\subseteq\mathcal D_n\) is strict.
Proof: extend the counterexample by a zero block
For larger dimensions, append a zero block. Any completely positive factorization of the enlarged matrix would give one for its leading principal five-by-five block, so the gap persists.