Symmetric matrix

A real matrix is symmetric when $A^T=A$. It represents a self-adjoint operator in the standard Euclidean inner product.

A real symmetric matrix has real eigenvalues, orthogonal eigenspaces for distinct eigenvalues, and an orthonormal basis of eigenvectors. From diagonalizability, apply the Gram-Schmidt process within each eigenspace; all its linear combinations stay in that eigenspace. This gives $A=Q\Lambda Q^T$ with an orthogonal matrix $Q$ and real diagonal $\Lambda$.

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.

A real symmetric matrix whose quadratic form is positive at every nonzero vector in the nonnegative orthant. By compactness of the nonnegative unit sphere, this is equivalent to a positive uniform lower bound $x^{T}Ax\ge \delta \Vert x{\Vert }_2^2$ on that orthant. It need not be a positive-definite 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 $Hw=(0,0,0,2,2{)}^T$.

Every real symmetric matrix has an orthonormal basis of real eigenvectors. Equivalently, it admits an orthogonal diagonalization $A=O^{T}DO$ with $D$ real diagonal and $O$ an orthogonal matrix.