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For a sufficiently smooth local map fixing the origin, a local invariant centre manifold is tangent to the generalized eigenspaces with eigenvalues of modulus one. Writing the map as $(u,v)\mapsto (F(u,v),G(u,v))$ and the manifold as $v=h(u)$ gives the invariance equation $h(F(u,h(u)))=G(u,h(u))$. Finite Taylor series coefficients can be found from this equation. If all transverse eigenvalues have modulus less than one, local asymptotic stability reduces to that of the map on the centre manifold.

For $x$ in the derived algebra of a complex matrix Lie algebra $L$, define $y$ by multiplication by $\overline{\lambda }$ on each generalized eigenspace $V_{\lambda }$ of $x$. Polynomial interpolation and adjoint compatibility of additive Jordan decomposition make $\mathrm{ad}y$ a polynomial in $\mathrm{ad}x$ with zero constant term, so $[y,L]\subseteq [L,L]$. Trace orthogonality of $[L,L]$ and $L$ forces the displayed sum to vanish. Thus $x$ is a nilpotent endomorphism, and the Engel theorem implies solvability. The auxiliary $y$ and the Jordan components are not required to lie in $L$.

A stationary iterative method for a linear system converges to its specified fixed point for every initial vector if and only if its iteration matrix satisfies $\rho (H)<1$, where the spectral radius is the largest modulus of an eigenvalue.^[ref]

For a diagonalizable matrix, write $H=V\Lambda V^{-1}$. Then $H^k=V{\Lambda }^k{V}^{-1}$, so each eigenvector coefficient tends to zero exactly when its eigenvalue has modulus less than one. For a general matrix, the Jordan normal form gives the same result: a Jordan block $J=\lambda I+K$ with a nilpotent matrix $K$ of size $m$ satisfiesIf $0<|\lambda |<1$, every coefficient tends to zero because exponential decay beats any fixed polynomial in $k$; if $\lambda =0$, the block vanishes once $k\ge m$. Conversely, an eigenvector with $|\lambda |\ge 1$ provides an initial error that cannot tend to zero. For a real matrix with complex eigenvalues, the same obstruction appears in the real invariant subspace spanned by the real and imaginary parts of an eigenvector.

The words "every initial vector" matter. For example, $H=\mathrm{diag}(3/2,1/2)$ fails the criterion, but $e^{(0)}=(0,1{)}^T$ still tends to zero. Here a generic initial error has a nonzero first coordinate and diverges; the exceptional errors lie on the second coordinate axis. With $H=\mathrm{diag}(-1,1/2)$, generic errors oscillate rather than diverge in norm, while the same exceptional axis still converges. Both are valid matrix splittings with $M=I$ and $A=I-H$. In general the convergent errors form the direct sum of the generalized eigenspaces for $|\lambda |<1$; if $\rho (H)\ge 1$, this is a proper linear subspace.

A small spectral radius describes eventual convergence, but does not imply that the Euclidean norm of the error decreases on every step for a non-normal matrix. An operator norm bound $\Vert H\Vert <1$ does give the stepwise estimate $\Vert e^{(k)}\Vert \le \Vert H{\Vert }^k\Vert e^{(0)}\Vert$. For a real symmetric matrix, the matrix 2-norm equals its spectral radius, making this particularly direct.

The Floquet growth rate measures the exponential trend of a periodic linear differential equation over many coefficient periods. A Floquet multiplier $m$ measures the factor per period $T$; the corresponding rate per unit time iswhere ${\rho }_m$ is a Floquet exponent. Taking the real part removes the ambiguity in the complex logarithm.

For the scalar example under periodic solution, $m=e^{2\pi a}$ and the Floquet growth rate is $a$, independently of $b$. Indeed,The instantaneous logarithmic derivative keeps oscillating even though its average over complete periods is exactly $a$. Sampling at the same phase in each period removes the modulation altogether: $y(nT)=m^{n}y(0)$.

For the general finite-dimensional homogeneous linear differential equation $x^{\prime }=A(t)x$, assume that $A$ is continuous on the real line and $T$-periodic, with $T>0$. Normalize its fundamental matrix by $\Phi (0)=I$ and set $M=\Phi (T)$. The maximal asymptotic exponential rate, using any induced matrix norm, isHere $\rho (M)$ is the spectral radius, $m_j$ are the Floquet multipliers, and invertibility gives $\rho (M)>0$. This concerns growth of the solution operator; it does not say that every initial condition attains the maximum.^[ref]

To prove the formula, write $t=nT+s$ with $0\le s<T$. Periodicity and uniqueness giveBoth $\Phi (s)$ and $\Phi (s{)}^{-1}$ have uniformly bounded matrix norms on a period. Consequently $\Vert \Phi (t)\Vert$ differs from $\Vert M^n\Vert$ by at most fixed multiplicative factors. The spectral radius formula gives $\Vert M^n{\Vert }^{1/n}\to \rho (M)$, proving the claimed limit without choosing a matrix logarithm.

For a particular nonzero initial condition $v$, decompose $v$ into the generalized eigenspaces of $M$ over the complex numbers. The rate of $\Phi (t)v$ is the largest $T^{-1}\log |m_j|$ among components that are nonzero. Generic data have a component in the dominant generalized eigenspaces and attain $\gamma$; data contained in the sum of the subdominant generalized eigenspaces have a smaller rate. For example, with $A=\mathrm{diag}(0.2,-0.3)$, the maximal rate is $0.2$, while $v=(0,1)$ decays at rate $-0.3$. These invariant subspaces describe initial data at phase zero. Their transported versions $E(t)=\Phi (t)E(0)$ can vary with time and satisfy $E(t+T)=E(t)$.

A negative Floquet growth rate allows temporary amplification. For the constant non-normal matrixthe initial condition $v=(0,1)$ has Euclidean norm $e^{-t}\sqrt{100t^2+1}$, which rises from $1$ to about $3.697$ at $t=1$ before eventually decaying. Its exponential rate is still $-1$: the Jordan block contributes a polynomial factor, whose logarithm divided by $t$ tends to zero. At a unit-modulus Floquet multiplier, such polynomial growth can even be unbounded with zero exponential rate; $A=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ gives $e^{tA}v=(t,1)$. Thus zero Floquet growth rate alone proves neither boundedness nor a periodic solution.

If $D$ is a derivation of a Lie algebra, then its generalized eigenspaces satisfyApply a sufficiently large power of $D-(\lambda +\mu )I$ and use the binomial theorem together with the derivation rule.

Every finite-dimensional complex Lie algebra representation of a nilpotent Lie algebra decomposes into invariant generalized weight spaces. The nilpotence assumption is essential: a general solvable action can have extensions between distinct weights.

Here is a proof. For $h,k\in H$, nilpotence gives $(\mathrm{ad}h{)}^{N}k=0$, hence $(\mathrm{ad}\rho (h){)}^N\rho (k)=0$. By the nilpotent commutator preserves generalized eigenspaces lemma, every generalized eigenspace of $\rho (h)$ is invariant under every $\rho (k)$. Split successively by the acting operators of a basis of $H$. Each resulting block is invariant under $H$, and each basis operator has only one eigenvalue on it. The Lie theorem triangularizes the action on each block; its diagonal entries therefore give one and the same linear functional $\lambda$ on every basis vector of that block. Every $\rho (h)-\lambda (h)I$ is strictly upper triangular there, proving that the block is $V^{\lambda }$. Distinct blocks have distinct weights, so this is a direct sum. On the zero generalized weight space, every acting operator is a nilpotent endomorphism, so the Engel theorem supplies a nonzero common annihilated vector whenever that space is nonzero.

Over an algebraically closed field, a linear operator $X$ has a unique decomposition into a commuting diagonalisable endomorphism $X_s$ and nilpotent endomorphism $X_n$. On its generalized eigenspace for $\lambda$, set $X_s=\lambda I$ and $X_n=X-\lambda I$. The Chinese remainder theorem makes both parts polynomials in $X$, so they preserve every $X$-invariant subspace. Uniqueness follows by restricting any other commuting decomposition to those generalized eigenspaces. Over a perfect field the semisimple part need only become diagonalizable after scalar extension.

For endomorphisms of a finite-dimensional complex vector space, suppose repeated commutation with $T$ annihilates $S$. Write the blocks of $S$ using the generalized eigenspaces of $T$. On $\mathrm{Hom}(V_{\lambda },V_{\mu })$, the operator $\mathrm{ad}T$ is $(\mu -\lambda )I$ plus a nilpotent endomorphism. If $\lambda \ne \mu$, it is invertible, so $(\mathrm{ad}T{)}^{N}S=0$ forces that off-diagonal block of $S$ to vanish. Thus $S$ preserves each generalized eigenspace. This allows a nilpotent Lie algebra representation to be split by generalized eigenvalues even when its acting operators do not commute.

At a hyperbolic saddle, the stable invariant subspace of $Df(0)$ is the sum of its generalized eigenspaces with negative real parts; the unstable subspace uses positive real parts. The stable manifold theorem gives invariant local manifolds tangent to these subspaces at the equilibrium, with the corresponding dimensions and the appropriate forward or backward convergence to the equilibrium.

Here the Jacobian is $\mathrm{diag}(1,-1)$, so the unstable tangent is the $x$ axis and the stable tangent the $y$ axis. For the stable graph write $x=h(y)=A{y}^2+B{y}^3+O(y^4)$. Invariance requires $h^{\prime }(y)(-y+3h(y{)}^2)=h+h^2+2hy+3y^2$. Comparing quadratic and cubic terms gives $-2A=A+3$ and $-3B=B+2A$, so $A=-1,B=1/2$.

For the unstable graph $y=g(x)=C{x}^2+D{x}^3+O(x^4)$, invariance is $g^{\prime }(x)(x+x^2+2xg+3g^2)=-g+3x^2$. Coefficients give $2C=3-C$ and $2C+3D=-D$, so $C=1,D=-1/2$. Thus the cubic graph expansion of a saddle invariant manifold isThese describe the local manifolds to cubic order, not exact global polynomial curves.

Work in finite dimension over $\mathbb{C}$; the characteristic-zero real version follows by complexification. The Cartan solvability criterion isIts semisimplicity consequence, also often called Cartan's criterion, is $\mathfrak{g}$ is semisimple if and only if its Killing form is nondegenerate. We prove both forms.

The auxiliary results used are as follows. The Lie theorem simultaneously triangularizes every finite-dimensional representation of a complex solvable Lie algebra. The Engel theorem says that a matrix Lie algebra consisting entirely of nilpotent endomorphisms has a common annihilated nonzero vector and a basis making all its matrices strictly upper triangular; in particular it is a nilpotent Lie algebra. Nilpotent means that the lower central series terminates, which implies that the derived series terminates as well. Finally, the Additive Jordan decomposition writes any complex endomorphism uniquely as $x=x_s+x_n$, where $x_s$ is diagonalizable, $x_n$ is nilpotent and they commute. On a generalized eigenspace $V_{\lambda }$, the two parts are $\lambda I$ and $x-\lambda I$, and both are polynomials in $x$. On endomorphisms, $\mathrm{ad}{x}_s$ is diagonalizable, $\mathrm{ad}{x}_n$ is nilpotent, and these are the Jordan parts of $\mathrm{ad}x$: the first assertion follows by splitting into maps between eigenspaces, and the second by expanding repeated commutators with the nilpotent $x_n$.

First prove the linear trace version: if $L\subseteq \mathfrak{gl}(V)$ andthen $L$ is solvable. Fix $x\in [L,L]$ and let $V=⨁V_{\lambda }$ be its generalized eigenspace decomposition. Define $y$ to act on $V_{\lambda }$ by the scalar $\overline{\lambda }$. We do not assume that $y$, $x_s$ or $x_n$ belongs to $L$.

On $\mathrm{Hom}(V_{\mu },V_{\lambda })$, $\mathrm{ad}x$ has the form $(\lambda -\mu )I+N$ with $N$ nilpotent, whereas $\mathrm{ad}y$ is multiplication by $\overline{\lambda -\mu }$. Choose a polynomial $q$ with $q(d)=\bar{d}$ for every occurring eigenvalue difference $d$, and with all its derivatives of positive order up to the relevant nilpotency index zero at $d$. Such a polynomial exists by Hermite interpolation, or the Chinese remainder theorem for the pairwise coprime powers of $(t-d)$. At $d=0$ its value is zero, so $q(0)=0$. Expanding $q(dI+N)$ now givesBecause $x\in L$ and $[L,L]$ is an ideal, every positive power of $\mathrm{ad}x$ sends $L$ into $[L,L]$. The zero constant term therefore gives $[y,L]\subseteq [L,L]$.

Express $x$ as a sum of commutators ${\sum }_j[a_j,b_j]$ with $a_j,b_j\in L$. Cyclicity of trace and the hypothesis implyBut on $V_{\lambda }$, $xy$ has trace $\dim (V_{\lambda })|\lambda {|}^2$, since the nilpotent part has trace zero. Henceforcing every eigenvalue of $x$ to vanish. Thus every member of the derived algebra $[L,L]$ is a nilpotent endomorphism. By the Engel theorem, $[L,L]$ is nilpotent and therefore solvable; its derived series terminates, and adjoining the initial term $L$ shows that $L$ is solvable.

Conversely, if $L$ is solvable, the Lie theorem makes it upper triangular. Commutators of upper-triangular matrices have zero diagonal, and multiplying such a commutator by an upper-triangular matrix still has zero diagonal. Therefore the trace condition holds. Apply this equivalence to $L=\mathrm{ad}\mathfrak{g}$. Its trace pairing is the Killing form, and solvability of its image is equivalent to solvability of $\mathfrak{g}$: the kernel is its abelian centre, so a terminating derived series in the quotient terminates after at most one more step in $\mathfrak{g}$. This proves the solvability criterion.

For the semisimplicity criterion, let $R$ be the radical of the Killing form. Invariance makes $R$ a Lie algebra ideal. For $x,y\in R$, their adjoint actions on $\mathfrak{g}/R$ vanish, so computing trace in a basis adapted to $R$ gives $B_R(x,y)=B_{\mathfrak{g}}(x,y)=0$. The solvability criterion applied to $R$ shows it is a solvable ideal. Thus a semisimple Lie algebra has $R=0$ and a nondegenerate Killing form.

Conversely, any nonzero solvable ideal has a last nonzero term $A$ in its derived series; this is an abelian ideal of $\mathfrak{g}$. For $x\in A$, $y\in \mathfrak{g}$, the map $\mathrm{ad}x\mathrm{ad}y$ sends $\mathfrak{g}$ into $A$ and vanishes on $A$, so has trace zero. Hence $A\subseteq R$. A nondegenerate Killing form therefore rules out nonzero solvable ideals, exactly the definition of semisimplicity.

For classification, nondegeneracy of the Killing form provides the duality and root-space pairings used to extract a reduced crystallographic root system from a semisimple Lie algebra. Classification then reduces to the finite Dynkin diagrams and reconstruction of the Lie algebra from its root data.

An $R$-point means a unital algebra homomorphism over a field $\alpha :k_q[x,y]\to R$, as in points of a noncommutative algebra. It is specified by $X=\alpha (x)$ and $Y=\alpha (y)$ satisfying $YX=qXY$. For $R=\mathbb{C}$ and $q$ not a root of unity, $q\ne 1$, so scalar commutativity gives $(1-q)XY=0$. Hence the scalar points are the union of the two coordinate axes:

For matrix-valued points of a noncommutative algebra, put $H={\mathbb{C}}^n$ and let $H_{\lambda }(X)=\ker (X-\lambda I{)}^n$ be a generalized eigenspace. From $XY=q^{-1}YX$,Therefore $Y{H}_{\lambda }(X)\subseteq H_{q^{-1}\lambda }(X)$, interpreting a missing eigenvalue as the zero space. In particular the linear subspaces $V_X={⨁}_{\lambda \ne 0}{H}_{\lambda }(X)$ and $H_0(X)$ are invariant under both matrices. If $\lambda \ne 0$, the numbers $\lambda ,q^{-1}\lambda ,\dots ,q^{-n}\lambda$ are distinct. A nonzero vector $Y^{n}v$ with $v\in H_{\lambda }(X)$ would force all $n+1$ intervening generalized eigenspaces to be nonzero, impossible in dimension $n$. Thus $Y^n{V}_X=0$: $Y$ is nilpotent on $V_X$, although $X$ is invertible there.

Apply the same argument with $X,Y$ exchanged, using $XY=q^{-1}YX$. The sum $V_Y$ of the nonzero $Y$ generalized eigenspaces is invariant under both matrices, and $X$ is nilpotent on it. Consequently $V_Y\subseteq H_0(X)$. Conversely, since $Y$ is nilpotent on $V_X$, its nonzero generalized eigenspaces all lie in $H_0(X)$. Decompose the restriction of $Y$ to that invariant space into its nonzero and zero generalized eigenspaces:The zero generalized eigenspace of $Y$ is $X$-invariant by the scaling identity, so $U$ is invariant under both matrices. On $U$, both matrices have only zero eigenvalues, hence are nilpotent matrices. This proves the spectral decomposition of a quantum-plane point:Here $V_X,V_Y$ mean sums of generalized eigenspaces, not merely ordinary eigenspaces; that distinction is necessary for non-diagonalizable points.