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Let $U\subseteq W\subseteq {\mathrm{End}}_{\mathbb{C}}(V)$ be vector subspaces and $M=\{T:[T,W]\subseteq U\}$. If $\alpha \in M$ and $\mathrm{tr}(\alpha \beta )=0$ for all $\beta \in M$, then $\alpha$ is a nilpotent endomorphism.

To prove it, let $\beta$ act by $\overline{\lambda }$ on the generalized eigenspace of $\alpha$ of eigenvalue $\lambda$. Polynomial interpolation on eigenvalue differences, together with adjoint compatibility of additive Jordan decomposition, expresses $\mathrm{ad}\beta$ as a polynomial in $\mathrm{ad}\alpha$ with zero constant term. Since $\mathrm{ad}\alpha$ maps $W$ into $U$ and preserves $U$, this gives $\beta \in M$. Taking the matrix trace on each generalized eigenspace yields $0={\sum }_{\lambda }\dim (V_{\lambda })|\lambda {|}^2$, so every eigenvalue is zero. This is the linear-algebra step behind the Cartan solvability criterion; $U,W$ need not be Lie subalgebras.

For a positive witness acceptance operator on dimension $D$, large trace powers approximate its largest eigenvalue. With an $m$-qubit QMA quantum witness and $d=2m+2$, the usual NO instances give $\mathrm{Tr}(F^d)<2^{-d}$ and YES instances give a value above $2^{-d}$. A closed product-index expansion computes the trace by recomputation with polynomial storage, avoiding logarithms and roots in a real-arithmetic implementation.

The normalized eigenfunction ${\psi }_0(x)=\mathrm{sech}x/\sqrt{2}$ has zero eigenvalue under the Sine-Gordon kink fluctuation operator. It is proportional to ${\partial }_x{\varphi }_K$ and comes from shifting the collective coordinate of the kink. Its frequency is zero, so it contributes no oscillator zero-point energy; it must be handled separately from a Gaussian functional determinant.

A nonconstant smooth periodic roll obeys $\mu ⟨T_0^{\prime 2}⟩-⟨T_0^{\prime \prime 2}⟩-⟨T_0^{\prime 4}⟩=0$. The orthogonal-roll perturbation raises the increasing-gradient functional for poorly conducting convection by the displayed amount. The variance is strictly positive: a nonconstant periodic function has a derivative that vanishes somewhere and is nonzero somewhere. The Hessian of $V$ therefore has a positive direction. Its self-adjoint linearized evolution operator has a positive Rayleigh quotient and hence a positive eigenvalue, proving transverse linear instability wherever a nonconstant roll exists.

Let $H_0=\mathrm{diag}(-{\omega }_1,0,-{\omega }_2)$ and $H_1=d_1(E_{12}+E_{21})+d_2(E_{23}+E_{32})$, with nonzero $d_j$, distinct positive ${\omega }_j$. The squared adjoint action of $-i{H}_0$ has different eigenvalues $-{\omega }_1^2,-{\omega }_2^2$ on the two adjacent skew-Hermitian couplings. Linear combinations of $-i{H}_1$ and its double commutator with $-i{H}_0$ isolate both couplings. Further commutators give their real antisymmetric partners, two traceless diagonal generators, and both $1$–$3$ generators, hence the special unitary Lie algebra $\mathfrak{su}(3)$. The drift has nonzero trace; subtracting its traceless part supplies $iI$, giving the Lie algebra of the unitary group $\mathfrak{u}(3)$ and unitary operator controllability under the usual freely timed real-control assumptions.

If $A$ is diagonalizable, the anticommutator map $B\mapsto AB+BA$ has eigenvalues $2{\lambda }_1$, ${\lambda }_1+{\lambda }_2$ twice, and $2{\lambda }_2$. Their product gives the displayed characteristic polynomial. The density of diagonalizable complex matrices and continuity of polynomial coefficients prove the formula also for defective matrices.

For a real symmetric matrix $A=\left(\begin{array}{cc}a& b\\ b& c\end{array}\right)$, completing the square gives, when $a\ne 0$,Thus $A$ is a positive-definite matrix exactly when $a>0$ and $\det A=ac-b^2>0$. For necessity, test $(1,0)$ first, then $(-b/a,1)$; for sufficiency, the displayed two positive squares give the result. These are the first two leading principal minors in Sylvester's criterion. A positive determinant alone is insufficient: $-I_2$ has determinant $1$ but is a negative-definite matrix.

The eigenvalues provide an equivalent test under the same real-symmetric hypothesis:The spectral theorem for real symmetric matrices supplies coordinates in an orthonormal basis in which $q={\lambda }_{+}{u}^2+{\lambda }_{-}{v}^2$. More generally, a complex Hermitian matrix has an orthonormal eigenbasis and $q(z)={\sum }_j{\lambda }_j|w_j{|}^2$, so it is a Hermitian positive-definite matrix exactly when every eigenvalue is positive. This is a short proof outline: test the eigenvectors for necessity, and sum the positive squared absolute values for sufficiency.^[ref]

For a real positive semidefinite matrix, the condition is instead $q(x,y)\ge 0$ for all vectors, equivalent here to $a\ge 0$, $c\ge 0$, and $ac-b^2\ge 0$. If $a>0$, the same completing the square proves this; if $a=0$, nonnegativity forces $b=0$ and leaves $c{y}^2$ with $c\ge 0$. Merely replacing the strict inequalities on the leading principal minors by nonstrict ones fails: $\mathrm{diag}(0,-1)$ has both leading principal minors zero yet gives $q(0,1)=-1$. An indefinite quadratic form takes both positive and negative values; its real symmetric matrix has eigenvalues of both signs.

For $u_t=u_x+u_y$ with equal spatial mesh sizes, the centered leapfrog recurrence has amplification polynomial of a multilevel finite difference scheme $G^2-2i\mu (\sin \xi +\sin \eta )G-1$. Uniform stability for arbitrary two-level starting perturbations holds exactly for $0<\mu <1/2$. At $\mu =1/2$ the phase pair $(\pi /2,\pi /2)$ gives $(G-i{)}^2$, and its companion matrix has a nontrivial Jordan block, producing growth proportional to the number of steps. For larger $\mu$ one polynomial root has modulus greater than one. Bounds on eigenvalue moduli alone therefore give a misleading non-strict endpoint.

A real $2\times 2$ Jacobian matrix has both eigenvalues in the open left half-plane exactly when its trace is negative and its determinant is positive. At zero trace the test no longer determines nonlinear stability.

Over an algebraically closed field, a finite-dimensional Lie algebra is nilpotent exactly when every two-dimensional Lie subalgebra is abelian. A nonnilpotent Adjoint representation matrix has a nonzero eigenvalue $\lambda$ with eigenvector $y$, so $[x,y]=\lambda y$ generates a nonabelian two-dimensional subalgebra. Conversely a subalgebra of a nilpotent Lie algebra is nilpotent, while a nonabelian two-dimensional algebra has a basis $u,v$ with $[u,v]=v$ and a nonnilpotent adjoint matrix. Algebraic closure is what guarantees the required eigenvector.

For $L=\left(\begin{array}{cc}{\lambda }_1& 0\\ 1& {\lambda }_2\end{array}\right)$ with distinct real negative eigenvalues, the matrix exponential is $A=\left(\begin{array}{cc}a& 0\\ b& d\end{array}\right)$, where $a=e^{{\lambda }_{1}t}$, $d=e^{{\lambda }_{2}t}$ and $b=(a-d)/({\lambda }_1-{\lambda }_2)$. The optimal energy amplification of a linear system isIts eigenvectors are nonorthogonal, and immediate energy growth occurs exactly when ${\lambda }_1{\lambda }_2<1/4$. If ${\lambda }_2>{\lambda }_1$, putting $\alpha =({\lambda }_2-{\lambda }_1{)}^{-1}$ gives $G(t)\sim (1+{\alpha }^2)e^{2{\lambda }_{2}t}$. The optimal asymptotic initial direction $(\alpha ,1)$ is an adjoint eigenvector, while the eventual state aligns with the right eigenvector $(0,1)$.

For two centered predictors with squared norms $n$ and correlation coefficient $\rho$, the ordinary least squares estimate of either partial slope has variance ${\sigma }^2/[n(1-{\rho }^2)]$. Compared with the same predictor alone and the same error variance, the inflation factor is $1/(1-{\rho }^2)$. The eigenvalues of the slope Gram matrix are $n(1+\rho )$ and $n(1-\rho )$.

For smooth positive $K$ on an interval with regular Dirichlet boundary conditions, there is no turning-point phase shift. Combining conjugate WKB approximation branches to vanish at the first endpoint makes the total phase at the second an integer multiple of $\pi$. The first correction is the integrated WKB phase correction for a slowly varying oscillator. For $K=x+1$ on $[0,1]$ the resulting eigenvalues are $4{\pi }^2{n}^2/9-3/16+O(n^{-2})$.

A uniaxial nematic order parameter has one distinguished nematic director and equal eigenvalues in its perpendicular plane. In the displayed normalization the eigenvalues are $\lambda ,-\lambda /2,-\lambda /2$, giving $\mathrm{Tr}{Q}^2=3{\lambda }^2/2$ and $\mathrm{Tr}{Q}^3=3{\lambda }^3/4$. Positive and negative $\lambda$ describe prolate and oblate orientational anisotropy, respectively; reversing the nematic director leaves the tensor unchanged.

An element $u$ of a ring is unipotent when $u-1$ is nilpotent. For a matrix, this says that every eigenvalue is one. A unipotent algebraic group consists of unipotent elements in a faithful linear representation.

A square matrix $U$ is unipotent when $U-I$ is nilpotent. Over an algebraically closed field, this is equivalent to all eigenvalues being one. If $j$ belongs to a nilpotent ideal of an operator algebra, $I+j$ is a unipotent matrix. This realizes the radical factor in the Levi decomposition of a quiver automorphism group.