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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.

The matrix 2-norm is the operator norm induced by the Euclidean norm:It equals the largest singular value of $A$ and is invariant under multiplication by orthogonal or unitary matrices.

The matrix norm induced by Euclidean vector norms is $\Vert A{\Vert }_2={\sup }_{x\ne 0}\Vert Ax{\Vert }_2/\Vert x{\Vert }_2$, equal to the largest singular value.

Write the spline interpolation operator as $P_{\mathbf{x}}=T{A}_{\mathbf{x}}^{-1}R$, where $R$ samples at the interpolation sites and $T$ synthesizes the normalized B-spline expansion. Sampling has operator norm at most one, and nonnegative B-splines with sum at most one give $\Vert T\Vert \le 1$. HenceHere the matrix norm is the maximum absolute row sum. Existence of the inverse matrix requires the usual ordered distinct interpolation sites and the Schoenberg–Whitney theorem support conditions, with endpoint values interpreted by their one-sided limits. If the displayed diagonal condition is read without the implicit ordering and distinctness, it is insufficient: all three sites equal to $1/2$ in the quadratic example give positive diagonal entries but three identical rows. The estimate applies whenever the interpolating linear map in the question is defined uniquely.

For the quadratic Bernstein basis, $N_1=(1-x{)}^2$, $N_2=2x(1-x)$ and $N_3=x^2$. Sampling and inverting giveThe three cardinal Lagrange interpolation polynomials areThe Lebesgue constant of interpolation gives the exact operator norm: the upper bound follows from $|Pg(x)|\le \Vert g{\Vert }_{\infty }{\sum }_i|{\ell }_i(x)|$. To attain it at a maximizing point, prescribe at the three sites the signs of the corresponding cardinal polynomials and extend those values by a continuous piecewise linear function of norm one.

On $0\le x\le 1/2$, the first two cardinal polynomials are nonnegative and the last is nonpositive. Since their sum is one, their absolute sum is $1-2{\ell }_2(x)=1+2x-4x^2$. Its maximum is $5/4$ at $x=1/4$. Reflection gives the same maximum at $3/4$ on the other half. In particular the data $(1,1,-1)$ produce $1+2x-4x^2$, attaining $5/4$. ThusThe coefficient estimate loses cancellation among the B-splines, which explains why it is not sharp for the resulting function norm.

Use the cross-product Lie algebra convention $X=\widehat{e_1}$, $Y=\widehat{e_2}$, $Z=\widehat{e_3}$, so $[X,Y]=Z$. Put $R_x(t)=\exp (tX)$ and $R_y(t)=\exp (tY)$, using the matrix exponential. The finite group commutator isHere $\epsilon \to 0$ and big O notation refers to any fixed matrix norm. Active rotation matrices act on column vectors from right to left: first rotate about $y$ by $-\epsilon$, then about $x$ by $-\epsilon$, then about $y$ by $+\epsilon$, and finally about $x$ by $+\epsilon$. Swapping $x$ and $y$ replaces the product by its inverse and changes the leading Lie bracket to $-Z$.

Writing $c=\cos \epsilon$ and $s=\sin \epsilon$, direct matrix multiplication givesFor $\epsilon =0.1$ radians, $C_{\epsilon }{e}_1\approx (0.999950208,0.009966462,0.000501230)$, whereas $e_1+{\epsilon }^{2}Z{e}_1=(1,0.01,0)$. Thus the residual is a positive rotation about $z$ to leading order; at finite $\epsilon$ it also has smaller components. The Lie bracket captures the leading effect of the four moves rather than their entire finite motion. See Etingof, Section 7.2, Proposition 7.11.

The left panel projects the actual four arcs onto the $yz$-plane for $\epsilon =0.25$; the start and endpoint are close but distinct. The right panel divides the residual coordinates by ${\epsilon }^2$, making the limit $(\Delta y/{\epsilon }^2,\Delta z/{\epsilon }^2)\to (1,0)$ visible. This is a numerical picture of $[X,Y]=Z$ in the SO(3) Lie algebra.