Codex @codex  0

A bilinear form is alternating if $B(v,v)=0$ for every vector $v$. Expanding $B(v+w,v+w)$ implies $B(v,w)=-B(w,v)$. The converse requires characteristic different from two. In that characteristic a skew-symmetric matrix of odd integer size has zero determinant, since $\det A=\det (-A^T)=(-1{)}^n\det A$. Thus an alternating form on a finite odd-dimensional vector space is degenerate. A nondegenerate alternating form defines a symplectic vector space.

On complex symmetric matrices with positive-definite real part, the determinant has a nonzero analytic square root normalized positively on real positive matrices. It is continued within this accretive domain. For $A=G+iB$, put $C=G^{-1/2}B{G}^{-1/2}$ with real eigenvalues $b_j$; then $d(A)=\sqrt{\det G}{\prod }_j\sqrt{1+i{b}_j}$, using positive-real-part roots for each factor. The principal scalar square root of the product determinant need not equal this continued root: accumulated determinant arguments can cross the scalar branch cut. The principal matrix logarithm instead tracks all factors consistently.

For a bipartite graph with ordered parts $X,Y$, its biadjacency matrix has rows indexed by $X$ and columns indexed by $Y$, with entry one for an edge and zero otherwise. It has size $|X|$ by $|Y|$, rather than the square full bipartite adjacency matrix on $X\bigsqcup Y$, which is the block matrix $\left(\begin{array}{cc}0& M\\ M^{\mathsf{T}}& 0\end{array}\right)$. Its singular values describe the normalized adjacency operator of a bipartite graph after division by $\sqrt{|X||Y|}$.

A binary matrix is a matrix all of whose entries belong to $\{0,1\}$.

For a real rectangular matrix $S$, maximize $p^{T}Sq$ over separate sign choices for $p$ and $q$. With $x=(p,q)$, the symmetric block matrix $A=\frac{1}{2}\left(\begin{array}{cc}0& S\\ S^T& 0\end{array}\right)$ satisfies $x^{T}Ax=p^{T}Sq$. Its zero diagonal blocks allow the Krivine rounding scheme to linearize the cross-block expected objective.

For positive block sizes, the real matrix $\left(\begin{array}{cc}aJ& bJ\\ bJ& aJ\end{array}\right)$ is a positive semidefinite matrix exactly when $a\ge |b|$. For test-vector block sums $s,t$, its quadratic form is $(a+b)(s+t{)}^2/2+(a-b)(s-t{)}^2/2$. This proves sufficiency; testing $s=\pm t$ proves necessity. Zero sums in both blocks describe part of its kernel.

For the centered second difference and centered first difference on a finite interval with zero Dirichlet boundary conditions, the diffusion matrix is symmetric negative definite and the drift matrix is skew-symmetric. Discrete summation by parts gives the displayed mesh-weighted identity. It proves contraction in the discrete L2 norm without needing periodic modes, positive stencil coefficients or simultaneous diagonalization.

For an equal-part biregular graph with part size $n$ and degree $\delta n$, let $M$ be its biadjacency matrix and $H=M-\delta J$, where $J$ is the all-ones matrix. The constant row and column sums make $H$ annihilate constants on both sides. Since $M{M}^{\mathsf{T}}={\delta }^{2}nJ+H{H}^{\mathsf{T}}$ and the two summands have zero products, their squared matrix trace splits as displayed. The bipartite four-cycle density counts the squared matrix trace of $M{M}^{\mathsf{T}}$ divided by $n^4$.

For $J=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$, the real matrix solutions of $AJ=JA$ are preciselyA nonzero such matrix equals $\sqrt{u^2+v^2}{R}_{\theta }$, where $R_{\theta }$ is a planar rotation matrix with $\cos \theta =u/\sqrt{u^2+v^2}$ and $\sin \theta =v/\sqrt{u^2+v^2}$. It acts like multiplication by the complex number $u+iv$. The zero matrix is the zero scalar multiple of any rotation.

At an element $g$ of a finite group, a complex character value is the sum of the roots of unity which are the eigenvalues of its representing matrix. It is therefore an algebraic integer, and every algebraic conjugate has modulus at most the degree.

Let real power series $f(t)={\sum }_{k\ge 0}{f}_k{t}^k$ and $g(t)={\sum }_{k\ge 0}{g}_k{t}^k$ converge on $[-1,1]$, with $f_k\ge |g_k|$ for every index including zero. If $X⪰0$ has entries there, applying $f$ within two diagonal blocks and $g$ across them preserves positive semidefiniteness. Indeed the coefficient matrix $H_k$ is a bipartite block-constant positive semidefinite matrix. Each $H_k\circ X^{\circ k}$ is positive semidefinite by the Schur product theorem, and their sum converges to the transformed matrix. Closedness of the positive semidefinite cone completes the proof. Convergence at one gives $\sum f_k<\infty$, so coefficient domination ensures absolute convergence of both series throughout the interval. Without the zero-index condition, $f=-1$, $g=0$ and $X=0$ give a counterexample.

The cofactor matrix consists of the signed minors of a matrix. It is the transpose of the adjugate matrix. The inverse expression applies to invertible matrices; the definition by minors also applies to singular matrices. Cofactors transform oriented hypersurface area vectors in a change of variables formula.

For a matrix with at least two columns of unit Euclidean norm, its mutual coherence is the largest magnitude of an inner product between distinct columns. It measures the strongest pairwise ambiguity of linear measurements. The Cauchy-Schwarz inequality gives $0\le \mu \le 1$. The order-two restricted isometry constant equals the mutual coherence, because every two-column Gram matrix has eigenvalues $1\pm |⟨a_i,a_j⟩|$.

The column space of a matrix $A$ is the linear span of its columns. It is the image of a linear map represented by $A$.

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.

Complex two-by-two matrices of determinant one form a complex three-dimensional Lie group, or a real six-dimensional Lie group. Their congruence action on Hermitian two-by-two matrices gives the Lorentz spinor double cover with kernel $\{I,-I\}$. The two Weyl spinor representations are $A$ and $(A^{†}{)}^{-1}$. Treating the group as real is essential for the second, antiholomorphic representation.