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Iterating an affine map gives $x_k=M^{k-1}{x}_1+{\sum }_{j=0}^{k-2}{M}^{j}c$. A fixed point exists exactly when $c$ is in the image of a linear map $I-M$. If the spectral radius of $M$ is below one, its matrix powers tend to zero and every orbit converges to the unique fixed point $(I-M{)}^{-1}c$.

An algebra homomorphism onto a semisimple Banach algebra is continuous. To see the spectral mechanism, suppose $a_n\to 0$ and $T(a_n)\to b$. For any $c\in B$, choose $u,v\in A$ with $T(u)=b$ and $T(v)=c$, and set $d=cb$, $d_n=T(v{a}_n)$. Apply the spectral-radius three-circle inequality to $T((1-z)v{a}_n+zvu)$. On the outer circle its norm tends uniformly to $\Vert d\Vert$; on the circle of radius $1/R$ its spectral radius is bounded by $(1+1/R)\Vert v{a}_n\Vert +\Vert vu\Vert /R$. Hence $r(d{)}^2\le \Vert d\Vert \Vert vu\Vert /R$ for every $R>1$, so $r(cb)=0$. The unit criterion for the Jacobson radical gives $b\in J(B)=0$, and the closed graph theorem finishes the proof.

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.

Let $A=D+L+L^T$ be a real symmetric positive-definite matrix, with $D$ diagonal and $L$ strictly lower triangular. Put $B=D/\omega +L$ and $C_{\omega }=I-B^{-1}A$ for $\omega \ne 0$. The matrix norm induced by an energy norm reduces the question to $T=A^{1/2}{C}_{\omega }{A}^{-1/2}$. Direct multiplication and $B+B^T-A=(2/\omega -1)D$ giveThe left side is positive semidefinite. The matrix following the scalar coefficient on the right is positive-definite, since $D$ has positive diagonal and $B^{-1}{A}^{1/2}$ is an invertible matrix. For $0<\omega <2$, every eigenvalue of $T^{T}T$ therefore lies in $[0,1)$. Hence its spectral radius is below one and $\Vert C_{\omega }{\Vert }_A=\Vert T{\Vert }_2<1$. Iteration errors consequently decay geometrically in the energy norm. The argument proves convergence from every initial vector, without requiring the original iteration matrix to be symmetric.

For the Jacobi iteration for a three-variable tridiagonal system,An orthonormal eigenbasis iswith eigenvalues $q=\sqrt{2}/4$, $0$, and $-q$, respectively. These are error modes: their shapes persist under the iteration matrix, while their coefficients change. Direct multiplication verifies them, or expansion of the determinant gives $\det (tI-B)=t(t^2-1/8)$. Thus $\rho (B)=q<1$ and convergence holds for every starting vector.

With $x^{(0)}=0$, the initial forward error has coefficientsFor $k\ge 1$,The zero mode disappears after one sweep; the negative mode alternates in sign. The Euclidean norm drops by exactly $q$ per sweep from $k=1$ onward. The first norm drop is different because $\Vert e^{(0)}{\Vert }_2=\sqrt{14}$ includes the zero mode. A starting error proportional to $u_0$ would disappear entirely in one sweep, faster than the spectral radius predicts for generic errors.

The curves are computed iterates and exact formulas for this linear system. Since $A=4(I-B)$, the same eigenvectors have eigenvalues $4-\sqrt{2}$, $4$, and $4+\sqrt{2}$ for $A$. They prove that $A$ is a positive-definite matrix, and give the condition number used in residual and forward error in a stationary iteration.

The spectral radius of the matrix exponential gives the lower bound through the spectral abscissa $s(A)$. Differentiating the squared Euclidean norm along $q^{\prime }=Aq$ and bounding its Hermitian quadratic form gives the upper bound through the Euclidean logarithmic norm ${\mu }_2(A)$. A non-normal matrix can have ${\mu }_2(A)>0$ despite $s(A)<0$, allowing transient growth before its ultimate decay.

For the Gauss-Seidel iteration for a two-variable positive-definite system, define the forward error by $e_x^{(k)}=x^{(k)}-x_{\ast }$ and $e_y^{(k)}=y^{(k)}-y_{\ast }$. Subtract the two exact solution equations from the two update equations:The fresh first coordinate passes the old second-coordinate error through two successive factors. The iteration matrix is thereforeIts eigenvalues are $0$ and $1/12$, as seen from the determinant. The zero eigenvalue explains why the old first-coordinate error disappears completely: the first row overwrites it. After one sweep, every remaining forward error lies on the eigenvector line spanned by $(-3,1{)}^T$.

For this zero start, $e_y^{(0)}=-7/11$, so the exact vectors for $k\ge 1$ areThe Euclidean norm of the forward error decreases by exactly $1/12$ per sweep from $k=1$ onward. The first decrease is different because the initial error has not yet reached that eigenvector line. Each row update moves horizontally or vertically in the coordinate plane; sweep endpoints then approach the solution along one straight line.

The linear-system residual is an independent way to check the calculation. Completing a sweep satisfies the second row exactly, but changing $y$ disturbs the first row:Thus a zero residual in the last updated row alone does not certify a solution; inspect the complete linear-system residual.

For the same matrix, the stale schedule of the Jacobi method gives the iteration matrixHere its eigenvalues are $\pm 1/\sqrt{12}$, whereas the fresh schedule gives $\rho (H)=\rho (B{)}^2$. This comparison belongs to this particular linear system; it is not a universal relation between the two methods' spectral radii.

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.

A strictly diagonally dominant matrix or a real symmetric positive-definite matrix guarantees convergence of the Gauss-Seidel method from every initial vector. These are sufficient conditions.^[ref]

For strict diagonal dominance, putInduct through a sweep. Old coordinates have absolute error at most $E$; previously updated coordinates have absolute error at most $qE\le E$. The component update and the triangle inequality therefore bound the new error by $qE$. This proves $\Vert e^{(k+1)}{\Vert }_{\infty }\le q\Vert e^{(k)}{\Vert }_{\infty }$, a maximum norm contraction.

For a symmetric positive-definite matrix, write $A=D+L+L^T$. The Householder-John theorem applies to $M=D+L$ and $N=-L^T$, because $M^T+N=D$ is a positive-definite matrix. Hence the iteration matrix has spectral radius less than one, which gives the convergence criterion for a stationary iteration.

There is also a geometric reason fresh updates help: the Gauss-Seidel method is exact coordinate descent on the convex quadratic functionHolding the other coordinates fixed, minimizing in coordinate $i$ gives the corresponding row update. A change ${\delta }_i$ lowers $F$ by exactly $a_{ii}{\delta }_i^2/2$. Each fresh coordinate is therefore the minimizer for the current vector, and a complete sweep lowers the quadratic form error unless every coordinate already solves its row. This explains the method's beauty without implying monotone decrease in every vector norm.

An invertible matrix and nonzero diagonal entries make the update well defined, but do not guarantee convergence. For the original linear systemthe Gauss-Seidel method updates $x^{(k+1)}=1-2y^{(k)}$ and $y^{(k+1)}=1-2x^{(k+1)}=-1+4y^{(k)}$. Starting at zero gives $(1,-1)$, then $(3,-5)$. The second-coordinate forward error is $e_y^{(k)}=-4^k/3$, and the iteration matrix $H=\left(\begin{array}{cc}0& -2\\ 0& 4\end{array}\right)$ has spectral radius $4$. The linear system has a unique solution while this particular stationary iterative method for a linear system diverges.

For a three-by-three matrix with diagonal entries one and every off-diagonal entry $\alpha$, the Jacobi method iteration matrix has eigenvalue $-2\alpha$ on the all-ones vector and eigenvalue $\alpha$ with multiplicity two on its orthogonal complement. Its spectral radius is $2|\alpha |$, so Jacobi iteration converges exactly when $|\alpha |<1/2$.

A strictly diagonally dominant matrix guarantees convergence of the Jacobi method for every initial vector.^[ref]

Indeed, using the maximum norm $\Vert v{\Vert }_{\infty }={\max }_i|v_i|$, the induced operator norm is the largest absolute row sum. The Jacobi method has $b_{ii}=0$ and $b_{ij}=-a_{ij}/a_{ii}$ for $j\ne i$, soThis gives a cheap sufficient check without computing any eigenvalues. For the Jacobi iteration for a three-variable tridiagonal system, the absolute row sums of $B$ are $1/4,1/2,1/4$, giving $\Vert B{\Vert }_{\infty }=1/2$. The true spectral radius $\sqrt{2}/4$ gives a sharper eventual rate.

Strict diagonal dominance is sufficient, not necessary: $A=\left(\begin{array}{cc}1& 2\\ 1/10& 1\end{array}\right)$ violates it in the first row, but its Jacobi method has eigenvalues $\pm 1/\sqrt{5}$ and converges. Conversely, being a positive-definite matrix alone is insufficient: the Jacobi convergence for a three-by-three equicorrelation matrix with $\alpha =3/4$ has positive eigenvalues $5/2,1/4,1/4$ for $A$, but spectral radius $3/2$ for its iteration matrix. The extra tridiagonal matrix structure in Jacobi convergence for a symmetric positive-definite tridiagonal matrix is therefore substantive.

The joint spectral radius measures the worst exponential growth under arbitrary products from a finite family of matrices. Submultiplicativity gives its existence and invariance under changes of the chosen norm. Every fixed-length product $B$ satisfies $\rho (B{)}^{1/p}\le \hat{\rho }(\mathcal{A})$, while the maximum norm of all length-$p$ products to the power $1/p$ is an upper bound. A repeated product gives the first inequality; grouping an arbitrary long word into blocks proves the second. The spectral radius of each individual matrix does not control growth of all switched products.

For the five-point Dirichlet Laplacian on an $n\times n$ square grid, a block-Jacobi splitting that solves each horizontal line exactly has blocks $\mathrm{tridiag}(-1,4,-1)$. Its tensor-product sine modes have angles ${\theta }_r=r\pi /(n+1)$ and the displayed iteration matrix eigenvalues. With $c=\cos (\pi /(n+1))$, its spectral radius is $c/(2-c)<1$, so it converges on every fixed grid. However the mode $r=1,s=n$ has multiplier $-c/(2-c)\to -1$. It is highly oscillatory between lines and is barely damped. The undamped method is therefore not a uniformly effective smoother for the multigrid method, even though it converges. Convergence and smoothing test different properties of the error spectrum.

For every finite-dimensional normal matrix $A$, the matrix 2-norm equals the spectral radius:This follows by unitary diagonalization of a normal matrix and unitary invariance of the matrix 2-norm.

The entry $K_{ij}$ is the expected number of newly infected type-$i$ individuals produced by a newly infected type-$j$ individual while the population is disease-free. In a compartmental linearization, $F$ describes new-infection production and $V$ transitions and removal. Provided the infection-lifetime subsystem is transient, its integrated occupation matrix is $V^{-1}$, yielding $K=F{V}^{-1}$. The basic reproduction number is its spectral radius. This construction is described in the original next-generation-matrix derivation.

First give the general definitions and radius facts. The spectrum of a bounded operator is $\sigma (A)=\{\lambda :A-\lambda I\text{ has no bounded inverse}\}$, and the point spectrum is $p\sigma (A)=\{\lambda :\ker (A-\lambda I)\ne 0\}$. If $|\lambda |>\Vert A\Vert$, the Neumann series inverts $I-A/\lambda$, hence also $A-\lambda I=-\lambda (I-A/\lambda )$. Thus $\sigma (A)$ is contained in the closed disc of radius $\Vert A\Vert$ and is bounded.

The spectral radius is $\nu (A)={\sup }_{\lambda \in \sigma (A)}|\lambda |$. The spectral radius formula isFor the left shift $R$, an eigenvector satisfies $a_{j+1}=\lambda a_j$, so $a_j={\lambda }^{j-1}{a}_1$. A nonzero such sequence is in ${\ell }_1$ exactly when $|\lambda |<1$. Since $\Vert R\Vert =1$, its spectrum is contained in the closed unit disc. It contains the open unit disc as point spectrum and therefore its closure by closedness of the spectrum. Hence

The spectrum of a bounded operator is $\sigma (T)=\{\lambda \in \mathbb{C}:T-\lambda I\text{ has no bounded inverse}\}$. The point spectrum consists of those $\lambda$ for which $\ker (T-\lambda I)\ne 0$, namely the eigenvalues. The spectral radius is $r(T)={\sup }_{\lambda \in \sigma (T)}|\lambda |$.

If $\Vert T\Vert <1$, the Neumann series $S={\sum }_{n=0}^{\infty }(-T{)}^n$ converges in the Banach space of bounded linear operators, since its term norms are bounded by a convergent geometric series. Multiplying the finite sums on either side by $I+T$ gives $I-(-T{)}^{N+1}$, which tends in operator norm to $I$. Therefore

Writing $\mathcal{B}(X)$ for the Banach space of bounded linear operators on $X$, the spectrum isHere $I$ is the identity operator. Since $X$ is a Banach space, the bounded inverse theorem makes this equivalent to failure of bijectivity of $\lambda I-T$.

The spectral radius isFor a nonzero complex Banach space, the spectrum of a bounded linear operator is nonempty and compact, so the supremum is a maximum. If the zero space is allowed, its empty spectrum is assigned spectral radius zero.

Start the Galton-Watson process with one ancestor, $Z_0=1$, and let every individual independently have a number of offspring distributed as $Z$. With independent copies $Z_{t,j}$,where an empty sum is zero. Extinction is the event that some generation is empty; zero is then absorbing. Write $q$ for its probability and $1-q$ for the survival probability.

Set $q_t=P(Z_t=0)$, with $q_0=0$. Conditional on the ancestor's number of children, extinction by generation $t+1$ requires independent extinction by generation $t$ in every child subtree. ThusThe events $\{Z_t=0\}$ increase, so $q_t\uparrow q$. The probability generating function is continuous on $[0,1]$, by domination of $s^Z$ by $1$, giving $q=f_Z(q)$. If $r\in [0,1]$ is any fixed point, then $q_0=0\le r$ and monotonicity of $f_Z$ inductively gives $q_t\le r$. Taking limits proves the Galton-Watson extinction fixed point characterization:In particular it is the smallest nonnegative fixed point wherever the generating series is defined, since it already lies in $[0,1]$.

Let $m=\mathbb{E}Z=f_Z^{\prime }(1-)$, allowing $m=\infty$. The probability generating function is increasing and convex, with $f_Z(1)=1$. If $m<1$, its tangent inequality gives $f_Z(s)\ge 1+m(s-1)>s$ for $s<1$, so $q=1$. If $m=1$ and $P(Z=1)\ne 1$, there is positive mass at both zero and some integer at least two; strict convexity gives $f_Z(s)>s$ for $s<1$, again $q=1$. The exceptional case $Z=1$ almost surely has one individual forever, so $q=0$ despite $m=1$.

If $m>1$, the left derivative at $1$ gives $f_Z(s)-s<0$ for $s<1$ sufficiently close to $1$. Since $f_Z(0)\ge 0$, continuity supplies a fixed point below $1$, possibly $0$. Thus survival has positive probability precisely when $m>1$, apart from deterministic one-child reproduction.

For the random bipartite graph, early exploration of a component alternates between the two vertex classes. Before many vertices are exposed, each discovered vertex has approximately $\mathrm{Poisson}(c)$ new neighbors in the other class, since $\mathrm{Bin}(n-O(1),c/n)$ tends to that distribution. The two-type mean offspring matrix iswith spectral radius $c$; equivalently, the mean growth over two generations is $c^2$. The branching-process comparison therefore predicts a giant component exactly whenFor $c\le 1$, the local process dies out almost surely, so no linear-sized component is expected; $c=1$ is the critical case, not a positive-survival exception. For $c>1$, the symmetric survival probability is $1-q>0$, where $q=e^{c(q-1)}$, predicting approximately $2n(1-q)$ vertices in the giant. This last passage from local survival to a unique giant is the usual random-graph exploration argument; the branching-process reasoning here supplies the requested threshold explanation rather than a proof of that global theorem.