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Every complex symmetric matrix admits a unitary congruence decomposition with a unitary matrix $U$ and nonnegative diagonal entries $s_j$ equal to its singular values. This differs from unitary similarity diagonalization. It reduces a multimode squeezed vacuum to independent oscillator squeezing factors.

Use the conic Carathéodory theorem in the real space of symmetric matrices, of dimension $N=n(n+1)/2$. A convergent sequence $B_{\ell }$ in the completely positive cone has padded factorizations $B_{\ell }={\sum }_{j=1}^N{x}_{\ell j}{x}_{\ell j}^T$ with nonnegative factors. Their total squared norms equal $\mathrm{tr}{B}_{\ell }$ and are uniformly bounded. A simultaneous convergent subsequence of the finite factor tuple gives $B={\sum }_j{x}_j{x}_j^T$ with $x_j\ge 0$. Thus the limit remains in the cone. The argument also bounds the required number of factors by $N$.

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.

Conic optimization using the completely positive cone. For a real symmetric matrix $Q$, the trace-normalized program $\min \{⟨Q,X{⟩}_F:X\in {\mathrm{CP}}_n,\mathrm{tr}X=1\}$ minimizes a weighted average of nonnegative-unit-vector Rayleigh quotients. The weights are the squared norms of the factors in $X={\sum }_j{x}_j{x}_j^T$. Hence a minimizing rank-one factor attains the same value as the original orthant minimum.

If the constant symmetric matrix $A$ satisfies $A\xi \cdot \xi \ge \theta |\xi {|}^2$, then every smooth function $u$ with compact support satisfiesIndeed, the Fourier transform of a derivative and the Plancherel theorem turn the squared norms into integrals with Fourier multipliers $|\xi {|}^4$ and $(A\xi \cdot \xi {)}^2$.

The real symmetric matrices that are copositive matrices form a closed convex cone:Each fixed $x$ gives a closed linear inequality in $A$. Their intersection is therefore closed and convex, and it is preserved by nonnegative scaling.

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.

Let $\alpha =\min \{x^{T}Qx:x\ge 0,\Vert x{\Vert }_2=1\}$ for a real symmetric matrix $Q$. Compactness gives attainment. Homogeneity shows $Q-\lambda I$ is a copositive matrix exactly when $\lambda \le \alpha$, provingIts conic program dual is the trace-normalized completely positive optimization problem. A minimizing vector gives $X=x{x}^T$, which certifies equality and dual attainment directly. In general $\alpha$ differs from the unrestricted smallest eigenvalue.

If a symmetric matrix $H$ has largest eigenvalue ${\lambda }_1$ separated from ${\lambda }_2$ by $\gamma >0$, and $P=v{v}^{\top }$ is its leading rank-one orthogonal projection matrix, then every unit vector $u$, with $Q=u{u}^{\top }$, satisfies the displayed inequality. Expanding $u$ in an orthonormal eigenbasis gives ${\lambda }_1-u^{\top }Hu\ge \gamma (1-|u^{\top }v{|}^2)$, while $\Vert P-Q{\Vert }_F^2=2(1-|u^{\top }v{|}^2)$. The analogous statement for a smallest eigenvalue follows by replacing $H$ by $-H$.

The Davis-Kahan theorem bounds the change of an invariant eigenspace of a symmetric matrix in terms of a matrix perturbation and an appropriate spectral gap. The rank-one eigenprojector perturbation bound below depends only on the unperturbed gap. Yu, Wang and Samworth's formulation gives population-gap bounds in greater generality.

The flux $F(p)=p/\sqrt{1+|p{|}^2}$ of the minimal surface equation for a graph has a positive definite symmetric matrix derivative. Its eigenvalues are $(1+|p{|}^2{)}^{-3/2}$ in the direction of $p$ and $(1+|p{|}^2{)}^{-1/2}$ orthogonally; at $p=0$ every eigenvalue is one. On $|p|\le M$ it satisfiesThus an averaged linearization of a nonlinear divergence-form equation is uniformly elliptic where the relevant gradients are bounded. Pointwise positivity alone does not supply a uniform bound on an unbounded gradient range.

For an integer $0\le k\le n$, the fantope is the convex hull of rank-$k$ orthogonal projection matrices in ${\mathbb{R}}^n$. Equivalently, it consists of real symmetric matrices with eigenvalues in $[0,1]$ and matrix trace $k$. To prove the equivalence, apply the spectral theorem for real symmetric matrices and express the eigenvalue vector as a convex combination of the zero-one extreme points of the capped simplex. The sum of the largest eigenvalues is its support function.

For complex matrices, the inner product $⟨A,B{⟩}_F=\mathrm{tr}(A^{\ast }B)$ is conjugate-linear in its first argument. On real symmetric matrices it becomes $\mathrm{tr}(AB)$. In particular $⟨A,x{x}^T{⟩}_F=x^{T}Ax$, identifying quadratic forms with linear tests on matrices.

For two normal matrices of the same size, a pairing of their eigenvalues has squared Euclidean discrepancy at most the squared Frobenius norm of their difference. For Hermitian matrices, the increasing real eigenvalue order gives such a pairing, because it minimizes squared discrepancy. For real symmetric matrices, the norm square is $\mathrm{Tr}(A-B{)}^2$.

For a real symmetric matrix $X$ and $1\le k\le n$,Indeed $Y=Q{Q}^T$ is a rank-$k$ orthogonal projection matrix in the fantope, and projecting onto the eigenvectors of the largest eigenvalues attains the support function maximum. Equivalently, the largest matrix trace of a compression to a $k$-dimensional vector subspace is the sum of the largest eigenvalues.

For circular planetary orbits with angular momenta $L_j=M_j\sqrt{G{M}_{\star }{a}_j}$, the inclination matrix has positive off-diagonal entries, zero row sums, and $L_j{B}_{jk}=L_k{B}_{kj}$. Positive diagonal symmetrization of a matrix with $W=\mathrm{diag}(L_j)$ makes $W^{1/2}B{W}^{-1/2}$ a symmetric matrix. Moreover,All eigenvalues are real and nonpositive. The zero eigenvalue is a common tilt of every orbital plane; nonzero modes describe relative nodal precession.

Let a real symmetric matrix $M$ have ordered eigenvalues ${\lambda }_1\ge {\lambda }_2\ge \cdots \ge {\lambda }_n$, with $n\ge 2$. If $A$ is obtained by deleting one matching row and column, its largest eigenvalue ${\mu }_1$ satisfiesThe Rayleigh quotient of $A$ is that of $M$ restricted to vectors whose deleted coordinate is zero. This gives the upper bound. Some nonzero linear combination of the top two orthonormal eigenvectors has that coordinate zero; normalizing it gives a quotient at least ${\lambda }_2$. The argument does not require distinct eigenvalues.

For a real quadratic form $Q(v)=v^{T}Av$ with $A$ a symmetric matrix, its stabilizer is $G_Q=\{g\in {\mathrm{GL}}_n(\mathbb{R}):g^{T}Ag=A\}$. Its Lie algebra consists exactly of $X$ satisfying $X^{T}A+AX=0$. Differentiation proves necessity; the matrix exponential proves sufficiency because $e^{t{X}^T}A{e}^{tX}=A$. The result holds even when the quadratic form is degenerate.

Two symmetric matrices are congruent when $C=P^{T}AP$ for an invertible matrix $P$. This is the transformation of a quadratic form under an invertible linear change of coordinates. It differs from matrix similarity, which uses $P^{-1}AP$. Completing the square diagonalizes a real quadratic form by matrix congruence.

A real matrix is nonnegative when every entry is nonnegative. This entrywise condition differs from being a positive semidefinite matrix: the symmetric matrix $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ is nonnegative but has eigenvalues $1$ and $-1$.