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If a diagonalizable matrix satisfies $B^k=0$ for a positive integer $k$, write $B=P\mathrm{diag}({\lambda }_j)P^{-1}$. Then every ${\lambda }_j^k=0$, so every eigenvalue is zero and $B=0$. Diagonalizability is essential: the matrix $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ is nonzero but its square is zero. More generally a nonzero nilpotent operator requires a nontrivial Jordan block.

For a complex $n\times n$ matrix $A$, $\mathrm{tr}(A^k)$ is the sum of the $k$th powers of its eigenvalues, counted with algebraic multiplicity. This does not require diagonalizability: a similar upper triangular matrix has the eigenvalues on its diagonal, and taking powers raises those diagonal entries to their corresponding powers. Newton's identities therefore recover its characteristic polynomial from the first $n$ such traces.

For the Jordan block $A=\left(\begin{array}{cc}2& 1\\ 0& 2\end{array}\right)$, $\mathrm{tr}A=4$ and $\mathrm{tr}(A^2)=8$. Hence $e_1=4$ and $e_2=(4^2-8)/2=4$, giving $\det (zI-A)=z^2-4z+4=(z-2{)}^2$. The repeated eigenvalue is recovered even though this matrix is not diagonalizable.^[ref]

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.

A matrix is defective when an eigenvalue has geometric multiplicity smaller than its algebraic multiplicity. Over a field in which the characteristic polynomial splits, this is precisely the obstruction to an eigenbasis: at least one Jordan block has size greater than one, and generalized eigenvectors are needed to complete a basis. The shear mapping in diagonalizable matrix is the simplest example.

A finite-dimensional linear operator is diagonalizable over a specified field when it has an eigenbasis. Equivalently, its matrix $A$ admitswhere $P$ is an invertible matrix whose columns are eigenvectors, and $D$ is a diagonal matrix of the corresponding eigenvalues. This change of basis reveals a linear map as independent one-dimensional scalings. The matrix may look coupled in the original coordinates, but each coordinate in the eigenbasis evolves on its own.

For the symmetric matrix $A=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ computed in eigenvector, takeThen $AP=PD$, hence $P^{-1}AP=D$. The orthonormal basis version is $Q=P/\sqrt{2}$, an orthogonal matrix satisfying $Q^{T}AQ=D$. Every real symmetric matrix has an orthonormal eigenbasis by the spectral theorem; a general diagonalizable matrix need not have orthogonal eigenvectors.

The immediate payoff is repeated application: adjacent factors $P^{-1}P$ cancel, giving $A^n=P{D}^n{P}^{-1}$. In this example,The component in $E_3$ grows while the component in $E_1$ stays fixed. After division by its Euclidean norm, this particular sequence approaches $(1,1{)}^T/\sqrt{2}$. This is the mechanism behind the power method; an initial vector entirely in $E_1$ would stay there and never find the dominant eigenvalue $3$.

Two different obstructions deserve comparison. The planar rotation $R$ in eigenvector is diagonalizable over $\mathbb{C}$ but not over $\mathbb{R}$, because its eigenvalues are $i$ and $-i$. By contrast, the shear mappinghas only $\lambda =1$, of algebraic multiplicity two, and $E_1=\mathrm{span}\{(1,0{)}^T\}$, of geometric multiplicity one. There is no second independent eigenvector, even over $\mathbb{C}$. This is a defective matrix, already a two-dimensional Jordan block. Its powers areThus even though both eigenvalues are one, repeated application can produce growth. Eigenvalues alone miss the coupling recorded by a Jordan block. A repeated eigenvalue is harmless when it has enough independent eigenvectors, as $I$ demonstrates. More generally, a matrix is diagonalizable exactly when its characteristic polynomial splits over the chosen field and every eigenvalue has equal geometric multiplicity and algebraic multiplicity. The proof outline is to join bases of the distinct eigenspaces; their direct sum fills the space exactly under these conditions.

A physical motivation is the separation of normal modes. Consider two equal masses $m$ between three identical springs of stiffness $k$, with the outer ends fixed. By Hooke's law and Newton's second law, the displacement vector $q=(q_1,q_2{)}^T$ obeysThe eigenvectors $(1,1{)}^T$ and $(1,-1{)}^T$ have eigenvalues $1$ and $3$: the masses move together in the first normal mode and oppositely in the second. Writing $q=a(t)(1,1{)}^T+b(t)(1,-1{)}^T$ separates the coupled equations into two harmonic oscillators,Their angular frequencies are $\sqrt{k/m}$ and $\sqrt{3k/m}$; the eigenvalues give squared angular frequencies after multiplication by $k/m$. This is the beauty of diagonalization: finding the right directions turns interacting motions into independent patterns. For a physical treatment, see Yen-Jie Lee's "Coupled Oscillators, Normal Modes" (MIT 8.03SC, 2016), and the local article normal modes of two equal masses between three springs.^[ref]

The only diagonalizable nilpotent matrix is the zero matrix. Indeed, if $A^p=0$ and $Av=\lambda v$ with $v\ne 0$, then ${\lambda }^p=0$, so $\lambda =0$. A basis of eigenvectors would make every diagonal entry zero and therefore $A=0$. Nonzero nilpotent matrices require nontrivial Jordan blocks, even if their characteristic polynomial has only one distinct root.

For distinct real or complex numbers ${\lambda }_j$ and positive integers $m_j$, the vector space spanned by $t^k{e}^{{\lambda }_{j}t}$, $0\le k<m_j$, is preserved by differentiation. In each ordered basis $(e^{{\lambda }_{j}t},t{e}^{{\lambda }_{j}t},\dots )$, its matrix has ${\lambda }_j$ on the diagonal and $1,2,\dots ,m_j-1$ just above it. This follows from the product rule. Rescaling to $t^k{e}^{{\lambda }_{j}t}/k!$ makes each block a Jordan block. In particular, differentiation is invertible on this vector space exactly when every ${\lambda }_j\ne 0$: the determinant is ${\prod }_j{\lambda }_j^{m_j}$.

This generation-based compartmental model replenishes the susceptible compartment by $B>0$ per step and transfers $\alpha S_t{I}_t$ individuals into the next infectious generation. Members of the current infectious compartment leave it after one step. In scaled variables $x=\alpha S$, $y=\alpha I$, $r=\alpha B$, the update is $x^{\prime }=x+r-xy$, $y^{\prime }=xy$.

Its only fixed point is $(S_{\ast },I_{\ast })=(1/\alpha ,B)$. There is no finite disease-free fixed point, since susceptible recruitment continues when $I=0$. The Jacobian matrix at the fixed point is $M=\left(\begin{array}{cc}1-r& -1\\ r& 1\end{array}\right)$, with determinant one and trace $2-r$. For $0<r<4$, its two distinct eigenvalues have modulus one, so the linearized iterates are bounded and oscillatory. At $r=4$, the repeated eigenvalue $-1$ has a nontrivial Jordan block, making generic linearized iterates grow linearly. For $r>4$, one eigenvalue has modulus greater than one.

Thus $0<r<4$ gives neutral linear stability, not asymptotic stability. Unit-modulus multipliers do not determine nonlinear stability. Recruitment can restore susceptibility above $1/\alpha$ after a wave, allowing repeated infection growth; a closed SIR model without replenishment has a different susceptible balance.

An invertible complex Jordan block $\lambda I+N$, with $N^s=0$, has logarithm $\ell I+{\sum }_{k=1}^{s-1}(-1{)}^{k+1}(N/\lambda {)}^k/k$, where $e^{\ell }=\lambda$. Applying this construction blockwise and conjugating back proves surjectivity of the matrix exponential on the complex general linear group. There is no globally single-valued continuous choice of logarithm.

Stability of the partial differential equation requires a bound uniform over refining meshes and time steps on every fixed physical time interval. This is distinct from demanding an all-time bound for one fixed matrix. A fixed matrix is power bounded exactly when all eigenvalues are in the closed unit disk and its unit-modulus Jordan blocks have size one. A time-step family can nevertheless have harmless polynomial physical-time growth: $G_{\tau }=I+\tau N$ with $N^2=0$ gives $G_{\tau }^n=I+n\tau N$, uniformly bounded for $n\tau \le T$ despite its nontrivial unit-modulus Jordan block. Mesh dependence and the scaling of the step must therefore be kept in an eigenvalue argument.

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 Floquet multiplier is an eigenvalue of a one-period linear evolution map. For a homogeneous linear differential equation with continuous coefficients of period $T>0$, this map is the monodromy matrix of a periodic linear system $N$. If $Nv=mv$ for a nonzero eigenvector $v$, the associated solution of a differential equation with $Y(0)=v$ satisfies $Y(t+T)=mY(t)$. Its exponential rate is $T^{-1}\log |m|$, the corresponding Floquet growth rate. A Floquet multiplier equal to one gives a $T$-periodic solution in that eigenvector direction; a nontrivial Jordan block at a unit-modulus Floquet multiplier can still produce polynomial growth.

For a nonlinear periodic orbit, the Floquet multipliers are eigenvalues of the one-period map of its variational equation. A nonconstant periodic orbit of a smooth autonomous differential equation has a Floquet multiplier equal to one in its tangent direction: differentiating the solution of a differential equation with respect to its starting phase gives a periodic solution of the variational equation. The derivative of a transverse Poincaré map removes this phase direction and retains the transverse Floquet multipliers. Transverse Floquet multipliers strictly inside the unit circle give local attraction, while one outside gives instability; a transverse Floquet multiplier on the unit circle requires further analysis. For a planar periodic orbit, the transverse Floquet multiplier is given by the divergence test for a planar periodic orbit.

For a continuous periodic coefficient matrix $A(t+L)=A(t)$, a fundamental matrix $X$ for $X^{\prime }=A(t)X$, normalized by $X(0)=I$, obeys $X(t+L)=X(t)X(L)$. The invertible matrix $X(L)$ has a complex matrix logarithm; set $B=L^{-1}\log X(L)$ and $P(t)=X(t)e^{-tB}$. Since $X(L)=e^{LB}$ commutes with $B$, direct substitution gives $P(t+L)=P(t)$ and proves the factorization. A real logarithm is not required for this complex representation.

The eigenvalues of $X(L)$ are Floquet multipliers. All solutions are bounded for forward time when every multiplier has modulus at most one and those on the unit circle have no nontrivial Jordan blocks. In a real undamped second-order scalar equation with no first-derivative term, the monodromy determinant is one. Distinct conjugate unit-modulus multipliers give bounded solutions; a nontrivial Jordan block at multiplier $1$ or $-1$ gives an unbounded second solution even when one periodic or antiperiodic solution exists.

A Hurwitz stable matrix has all eigenvalues strictly in the open left half-plane. For $\dot{x}=Ax$, this is equivalent to $e^{At}x\to 0$ for every initial vector. Jordan blocks contribute polynomial factors times $e^{\lambda t}$, which decay exactly when their eigenvalues have negative real part. Translating an affine equilibrium reduces its attractivity to this same condition.

A Jordan chain of a linear operator $T$ for an eigenvalue $\lambda$ is a list of nonzero vectors satisfying $(T-\lambda I)v_1=0$ and $(T-\lambda I)v_j=v_{j-1}$ for $2\le j\le r$. These vectors are linearly independent: in a vanishing linear combination, applying the largest relevant power of $T-\lambda I$ isolates the last coefficient times $v_1$, then descending induction removes all coefficients. On their span the operator has one Jordan block. This is different from a generalized eigenfunction that solves $(T-\lambda I)u=0$ outside the original function space.

For a nilpotent linear map with minimal polynomial $x^m$, set $K_i=\ker T^i$ and $d_i=\dim K_i$, with $d_0=0$. ThenFor strictness, choose $w$ with $T^{m-1}w\ne 0$. The vector $T^{m-1-i}w$ belongs to $K_{i+1}\setminus K_i$ for $0\le i<m$. For the second inequality, the rank-nullity theorem givesThe first image is contained in the second because $T(K_{i+1})\subseteq K_i$.

If the Nilpotent Jordan block sizes are $s_1,\dots ,s_r$, thenThus successive increments count the Jordan blocks reaching a given size, and $(d_i-d_{i-1})-(d_{i+1}-d_i)$ counts blocks of size exactly $i$, with $d_{m+1}=d_m$. The nullities determine the entire Jordan normal form. Conversely any positive nonincreasing integer increments summing to $\dim V$ are realized by taking the corresponding numbers of each Nilpotent Jordan block.

If $(M-\lambda I)e=0$ and $(M-\lambda I)v=e$, the vector $v$ is a generalized eigenvector and $e^{\lambda t}(v+te)$ solves $\dot{x}=Mx$. The polynomial factor distinguishes a nontrivial Jordan block from an ordinary repeated eigenvalue with a full eigenvector basis.

If a diagonalizable matrix has distinct eigenvalues $r,s$, its spectral projections of a diagonalizable matrix are $(A-sI)/(r-s)$ and $(rI-A)/(r-s)$. Raising the diagonal form to a nonnegative integer power gives the displayed formula. Equivalently, interpolate the scalar function $t^n$ by a degree-one polynomial at $r,s$, then apply it to $A$. The diagonalizability assumption matters: knowledge of two eigenvalues alone does not exclude nontrivial Jordan blocks.

For a square matrix over a field, the minimal polynomial $m_A$ is the unique monic polynomial of least degree satisfying $m_A(A)=0$. The Cayley-Hamilton theorem guarantees at least one such polynomial: the characteristic polynomial. The minimal polynomial divides every polynomial that annihilates $A$, so reducing matrix powers modulo $m_A$ can need fewer coefficients than reducing modulo the characteristic polynomial.

For example, $2I_2$ has minimal polynomial $t-2$, whereas the Jordan block $\left(\begin{array}{cc}2& 1\\ 0& 2\end{array}\right)$ has minimal polynomial $(t-2{)}^2$. Both have the same characteristic polynomial. Their different matrix powers are illustrated in repeated eigenvalues in the Cayley-Hamilton theorem.

For the homogeneous linear differential equation $\dot{Y}=K(t)Y$ with continuous coefficient matrix of period $P>0$, the monodromy matrix of a periodic linear system is the one-period map from $Y(0)$ to $Y(P)$. If $X$ is any fundamental matrix,Its eigenvalues are the Floquet multipliers. Multipliers are nonzero, since a fundamental matrix is invertible. Changing the starting phase conjugates $N$, so the Floquet multipliers and Jordan blocks are independent of that choice.^[ref]

This turns a continuous-time problem into repeated matrix multiplication:If $Nv=mv$, the associated solution of a differential equation obeys $Y(t+P)=mY(t)$ at every time, not merely at the sampled times. In particular, $Nv=v$ gives a nonzero $P$-periodic solution, and every nonzero $P$-periodic solution arises this way. If $Nv=-v$, the state changes sign after one period and repeats after two. A unit-modulus Floquet multiplier different from $1$ need not give repetition after one period, and a generalized eigenvector in a nontrivial Jordan block need not even give bounded motion.

For the scalar example under periodic solution, $N$ is the one-by-one matrix $(e^{2\pi a})$. Numerically solving a matrix problem uses the same idea: integrate one initial value problem for each column of $X(0)=I$ through one period, then compute the eigenvalues of $X(P)$. This is useful when an explicit solution of a differential equation is unavailable: one period determines the Floquet growth rate over arbitrarily many periods.

For piecewise constant coefficients, which are also covered by the one-period construction, $N$ is the time-ordered product of the segment matrix exponentials, with the earliest segment on the right. For example, take period $P=2$ and alternateSince $K_1^2=K_2^2=0$,Replacing the coefficient by its mean would instead give $e^{K_1+K_2}=\left(\begin{array}{cc}\cosh 1& \sinh 1\\ \sinh 1& \cosh 1\end{array}\right)$ and the incorrect rate $0.5$. The matrices do not commute. The scalar mean-coefficient formula therefore cannot simply be transferred to matrix systems. Nevertheless the Liouville formula for a fundamental matrix gives the correct determinant even in the noncommuting case:In the example, this is $1=m_{+}{m}_{-}$; it fixes the product of the Floquet multipliers, rather than either multiplier separately.

In Sur les équations différentielles linéaires à coefficients périodiques (1883), Gaston Floquet calls a function satisfying $f(t+P)=mf(t)$ periodic of the second kind, and distinguishes the genuinely periodic function obtained when $m=1$. This historical terminology captures precisely the difference between repetition of the coefficient and repetition of the state.^[ref]