Nilpotent chains and Jordan canonical form

The theorem and its prerequisites

Every nonconstant polynomial $p\in\mathbb C[z]$ has a complex root; consequently every degree-$n$ complex polynomial is a product of $n$ linear factors, counted with multiplicity. We prove this conclusion through Goursat, the Cauchy estimate and Liouville, with the prerequisites given below.

Holomorphic means complex differentiable at every point of an open set; entire means holomorphic on all of $\mathbb C$. The elementary foundations used below are real one-variable integration and its chain rule, continuity and differentiation of polynomials, the complex modulus laws, and compactness in the Euclidean plane (including shrinking compact triangles, uniform continuity and a Lebesgue number for a compact cover). These real-analysis and topology foundations are explicitly prerequisites. No complex polynomial-root theorem, maximum-modulus theorem, analytic power-series representation, identity theorem, residue theorem or argument principle is assumed.

Complex differentiability implies continuity, since $u(z+h)-u(z)=h[(u(z+h)-u(z))/h]\to0$. Expanding a product difference quotient gives $(uv)'=u'v+uv'$. Where $v(z)\ne0$,

Thus products and nonvanishing quotients of holomorphic functions are holomorphic. In particular $f(z)/(z-a)$ is holomorphic away from $a$, and a root-free polynomial has an entire reciprocal. These facts use difference quotients, not a polynomial-root or regularity theorem.

1. An integral records both a path and its orientation

A piecewise $C^1$ curve is a continuous map $\gamma:[a,b]\to\mathbb C$ that is continuously differentiable on finitely many subintervals, with one-sided derivatives at the joins. It is closed when $\gamma(a)=\gamma(b)$. Its length is $\ell(\gamma)=\int_a^b|\gamma'(t)|\,dt$. A contour integral of continuous $f$ along it is

with real and imaginary parts integrated separately. The differential $dz$ includes the direction of motion; replacing it by $|dz|$ would define a different integral.

A piecewise $C^1$ increasing change of parameter preserving endpoints leaves the integral unchanged by substitution. Reversal changes its sign. Concatenating two compatible curves adds their integrals. These statements also allow zero-speed pieces. They do not require the curve to be injective.

Estimation lemma. If $|f|\le M$ on the curve, then

Indeed, the modulus of a real-parameter integral is at most the integral of the modulus, and $|f(\gamma)\gamma'|\le M|\gamma'|$. This simple inequality is the main tool for showing that auxiliary arcs contribute nothing in a limit.

Worked example. On the positively oriented circle $z=a+re^{it}$, $0\le t\le2\pi$, with $r>0$,

For $n=-1$ the integrand times $dz$ is $i\,dt$. Otherwise it is $ir^{n+1}e^{i(n+1)t}\,dt$, whose integral vanishes. The answer for $n=-1$ is independent of radius, but changes sign if orientation is reversed.

2. Primitives turn path integrals into endpoint differences

A primitive of $f$ on an open set $U$ is a holomorphic function $F$ there with $F'=f$. The real chain rule applied on each smooth piece gives

for every contour in $U$. Hence a function with a primitive has zero integral around every closed contour in $U$.

Conversely, on a domain $U$, suppose continuous $f$ has zero integral around every closed contour. Open connected subsets of the plane are polygonally connected. Fix $a\in U$ and define $F(z)$ as the integral from $a$ to $z$ along any polygonal path in $U$. Closed-path cancellation makes it path independent. For sufficiently small $h$, the final straight segment from $z$ to $z+h$ stays in $U$, and

by continuity. Thus $F'=f$. The same construction works if polygonal closed paths alone have zero integrals; the resulting primitive then handles all contours.

3. Goursat's triangle argument: no continuity of the derivative assumed

Theorem. If $f$ is holomorphic on an open set containing a closed triangle $T$, then its integral around the positively oriented boundary of $T$ is zero.

Proof. Subdivide $T$ by joining side midpoints into four triangles. Interior edge integrals cancel, so the outer integral $I$ is the sum of four child integrals. At least one child has integral modulus at least $|I|/4$. Repeat inside such a child, obtaining nested closed triangles $T_n$, diameter $D/2^n$, perimeter $P/2^n$, and

Compactness and shrinking diameters give one common point $z_0$. Differentiability at $z_0$ says

The constant and linear terms have primitives, so contribute zero on the triangle boundary. For any $\eta>0$, eventually $|\varepsilon|\le\eta$ throughout $T_n$. Since $z_0\in T_n$, the estimation lemma gives

Thus $|I|\le\eta DP$. As $\eta$ is arbitrary, $I=0$. No continuity of $f'$ was used. ∎

Corollary: a primitive on each disk. If $f$ is holomorphic on a disk $D$, fix its centre $a$ and set $F(z)=\int_{[a,z]}f$. For $z,z+h\in D$ sufficiently close, the triangle with vertices $a,z,z+h$ lies in $D$. The triangle theorem gives $F(z+h)-F(z)=\int_{[z,z+h]}f$. The limit calculation of §2 yields $F'=f$. Consequently all closed-contour integrals in the disk vanish. The same argument works on any open convex set.

4. Homotopy and what “no hole” actually means

A fixed-endpoint homotopy between paths $\gamma_0,\gamma_1$ is a continuous map $H:[0,1]^2\to U$ with $H(s,0)=\gamma_0(s)$, $H(s,1)=\gamma_1(s)$, and $H(0,t),H(1,t)$ constant in $t$. It deforms one path into the other without leaving $U$. A domain is simply connected if every closed path can be contracted to its basepoint by such a homotopy.

Homotopy invariance. If $f$ is holomorphic on $U$, and two piecewise $C^1$ paths in $U$ are fixed-endpoint homotopic there, their integrals agree.

Proof using only local primitives. The compact image of $H$ is covered by finitely many disks on which $f$ has primitives. By the Lebesgue-number property and uniform continuity, subdivide the parameter square into a fine rectangular grid so that the image of each closed cell lies in one such disk. Join the images of adjacent grid vertices by straight segments. For each cell these segments lie in its disk because disks are convex. Its polygonal boundary integral is zero by its primitive. Sum over cells: internal segments cancel with their reversals. On the bottom and top edges, each original path subarc and its replacing straight segment lie in the same disk and have the same endpoints, hence the same integral by the local primitive. The side segments vanish because endpoints were fixed. What remains is the bottom integral minus the top integral. This proves the claim even if the homotopy's intermediate paths are not differentiable. ∎

The open set for the circular deformation

In the Cauchy proof, let $U$ be the original open holomorphy neighbourhood of $\overline{D(c,R)}$, and let $d=|a-c|<R$. Choose $0<\varepsilon<R-d$. The correct holomorphy set for $g(z)=f(z)/(z-a)$ is $U\setminus\{a\}$, including the outer circle; the open disk $D(c,R)\setminus\{a\}$ alone would exclude that circle. This open set includes the boundary circle, and the following inequalities justify the deformation inside it.

Put $c_t=(1-t)c+ta$, $r_t=(1-t)R+t\varepsilon$, and $H(s,t)=c_t+r_te^{2\pi i s}$. For $0\le s,t\le1$,

Thus the whole compact homotopy image lies in $U\setminus\{a\}$. The grid/local-primitive proof works for this free homotopy of closed loops: $H(0,t)=H(1,t)$, so the two polygonal side tracks are identical, have opposite boundary orientations and cancel. They need not be constant. The top and bottom circles are piecewise $C^1$; no differentiability of the intervening homotopy is needed. This supplies exactly the domain and moving-basepoint step used next.

1. A boundary integral determines an interior value

For $c\in\mathbb C$ and $R>0$, orient the circle $C:|z-c|=R$ counterclockwise. Suppose $f$ is holomorphic on an open neighbourhood of the closed disk $\overline{D(c,R)}$. For every $a$ with $|a-c|<R$,

The denominator is never zero on $C$. The formula is an interior-value theorem; it does not assert this value for an exterior point or a point on the contour.

Proof. Choose $\varepsilon>0$ with $|a-c|+\varepsilon<R$. The outer circle deforms to the small circle $C_\varepsilon:|z-a|=\varepsilon$ without passing through $a$: use centres $(1-t)c+ta$ and radii $(1-t)R+t\varepsilon$. Each moving circle encloses $a$ and stays in the original disk. Apply the grid/local-primitive proof of C02 to $f(z)/(z-a)$ on the punctured disk. Although the moving basepoint is not fixed, its two side tracks are identical and cancel, so closed-loop integrals are unchanged. Hence

The final term has modulus at most $2\pi\sup_{|z-a|=\varepsilon}|f(z)-f(a)|$, which tends to zero by continuity at $a$. This proves the formula. ∎

This proof does not presume that $(f(z)-f(a))/(z-a)$ has a holomorphic extension at $a$; that fact is a consequence of the regularity obtained below.

The first derivative, with a uniform bound

Only the first Cauchy derivative estimate is needed for Liouville. For an interior point $a$ of a fixed circle $C$, set $\delta=\min_{z\in C}|z-a|>0$. When $0<|h|<\delta/2$, subtraction of the two Cauchy formulas gives the difference quotient kernel

Its difference from $1/(z-a)^2$ is

If $M=\max_C|f|$ and $L=\ell(C)$, the estimation lemma bounds the error after division by $2\pi i$ and integration by $ML|h|/(\pi\delta^3)$, which tends to zero. Hence $f'(a)=(2\pi i)^{-1}\int_C f(z)/(z-a)^2\,dz$. For a circle of radius $r$ centred at $a$, this gives $|f'(a)|\le M_r/r$. Repeating the same argument gives the higher-order formula below; Liouville uses only the first-order estimate.

2. All higher derivatives and their bounds

For $a$ in the open disk the distance from $a$ to the contour is positive. On a sufficiently small disk around $a$, difference quotients of the kernel $1/(z-a)$ converge uniformly in $z\in C$ to $1/(z-a)^2$. The estimation lemma permits passing that limit under the contour integral. Repeating this argument yields

The uniform kernel argument proves the existence of these derivatives; it does not assume $f$ was initially $C^\infty$. The same formula shows they are continuous locally because the denominators stay uniformly separated from zero.

On a circle centred at $a$ of radius $r$, put $M_r=\max_{|z-a|=r}|f(z)|$. Length estimation gives the Cauchy estimates

The radius must belong to a disk on which the function is holomorphic up to a neighbourhood of its boundary. Taking a larger radius through a singularity is invalid.

Liouville and the polynomial-root theorem

For an entire function, the preceding estimate is available at every radius. Applying it to a bounded entire function gives Liouville; applying Liouville to a polynomial reciprocal proves root existence.

Liouville's theorem. A bounded entire function is constant. If $|f|\le M$ on $\mathbb C$, apply the $n=1$ estimate on a disk of arbitrary radius $r$ about any $a$: $|f'(a)|\le M/r$. Sending $r\to\infty$ gives $f'=0$, so the function is constant on the connected plane.

Fundamental theorem of algebra. Every nonconstant complex polynomial has a complex root. If $p$ had no root, $1/p$ would be entire. Its leading term implies $|p(z)|\to\infty$ as $|z|\to\infty$, so $1/p$ is bounded outside a large disk; continuity gives boundedness on the remaining compact disk. Liouville makes $1/p$ constant, contradicting the degree of $p$. Repeated division by linear factors then factors every degree-$n$ polynomial into exactly $n$ linear factors counted with multiplicity.

Elementary details of the polynomial contradiction

For completeness, a function $F$ with $F'=0$ on $\mathbb C$ is constant: along $\gamma(t)=a+t(b-a)$ the real and imaginary parts of $F(\gamma(t))$ have derivative zero, so the real mean-value theorem gives $F(a)=F(b)$.

Write $p(z)=a_nz^n+\sum_{k<n}a_kz^k$ with $n\ge1$, $a_n\ne0$, and $S=\sum_{k<n}|a_k|$. For $|z|\ge R=\max(1,2S/|a_n|)$,

If $p$ had no root, its reciprocal would be continuous everywhere and holomorphic, since

The displayed growth bound makes $1/p$ bounded outside the radius-$R$ disk. It is bounded on the remaining closed disk by compactness and continuity. Liouville makes $1/p$ a constant, which is nonzero, forcing $p$ to be constant, a contradiction. Thus a root $\alpha$ exists.

Finally $z^k-\alpha^k=(z-\alpha)\sum_{j=0}^{k-1}z^{k-1-j}\alpha^j$ gives $p(z)-p(\alpha)=(z-\alpha)q(z)$, where $q$ has degree $n-1$ and leading coefficient $a_n$. Apply the root theorem to $q$ if it is nonconstant and induct on degree. After exactly $n$ divisions the remaining factor is the nonzero constant $a_n$. Degree and the product identity give exactly $n$ linear factors with multiplicity. This also proves the complex splitting assertion needed by finite-dimensional spectral or Jordan arguments; it does not assert splitting over an arbitrary real or other ground field.

Realification and a complete submultiplicative matrix norm

Real inner products, their norms and the needed inequalities

An inner product on a real vector space is a map $\langle u,v\rangle\in\mathbb R$ that is symmetric, linear in each argument, and positive definite: $\langle v,v\rangle>0$ for $v\ne0$. Linearity gives $\langle0,v\rangle=0$. Define $\|v\|=\sqrt{\langle v,v\rangle}$. It is zero exactly at zero and satisfies $\|cv\|=|c|\|v\|$. The standard example on $\mathbb R^m$ is $\langle u,v\rangle=\sum_{i=1}^mu_iv_i$. Vectors are orthogonal when their inner product is zero, and a list is orthonormal when distinct members are orthogonal and each member has norm one.

If $v\ne0$, insert $t=\langle u,v\rangle/\|v\|^2$ into the nonnegative square $\|u-tv\|^2$. Expanding by bilinearity gives $\|u\|^2-\langle u,v\rangle^2/\|v\|^2\ge0$, hence $|\langle u,v\rangle|\le\|u\|\|v\|$. If $v=0$ the same inequality holds directly. Equality for $v\ne0$ means $\|u-tv\|=0$, equivalently $u=tv$; the zero-vector cases also give equality. Finally expand $\|u+v\|^2$ and apply the bound to its cross term to obtain $\|u+v\|^2\le(\|u\|+\|v\|)^2$. Both sides to be square-rooted are nonnegative, so the triangle inequality follows. Thus this is a norm, and normalization of any nonzero vector divides by a positive real number. No complex bilinear convention is being used.

Why a norm on a finite real coordinate space is complete

Fix an integer $n\ge1$ and the Euclidean norm on $\mathbb R^n$. A set is bounded if it lies in some finite-radius ball, closed if its complement is open, and compact if every cover by open subsets of $\mathbb R^n$ has a finite subcover. We use real completeness, including the nested closed interval property.

Heine–Borel in finite-dimensional Euclidean space

Theorem. A set $K\subseteq\mathbb R^n$ is compact if and only if it is closed and bounded. The empty set satisfies all three properties.

Compact implies bounded. The open balls $B(0,m)$, $m\ge1$ an integer, cover $K$. A finite subcover puts $K$ inside the largest of these balls.

Compact implies closed. Suppose $K\ne\varnothing$ and $y\notin K$. The balls $B(x,\|x-y\|/2)$, $x\in K$, cover $K$. Choose a finite subcover with centers $x_1,\ldots,x_r$ and set $\delta=\min_i\|x_i-y\|/2>0$. If $z$ lies in the ball centered at $x_i$, the triangle inequality gives $\|z-y\|>\|x_i-y\|/2\ge\delta$. Hence $B(y,\delta)$ misses $K$, so the complement is open.

Closed and bounded implies compact. Let nonempty $K$ be closed and bounded. Enclose it in a closed cube $Q_0=[-M,M]^n$ with $M>0$. Suppose an open cover of $K$ has no finite subcover. Divide each coordinate interval into two closed halves. The resulting $2^n$ closed subcubes cover $Q_0$. At least one subcube has intersection with $K$ that is not finitely coverable by the given family; otherwise the finite union of the finite subcovers would cover $K$. Select such a subcube and repeat. This produces nested closed cubes $Q_j$ with side length $2M/2^j$ and $K\cap Q_j$ not finitely coverable, hence nonempty.

For each coordinate, the nested interval property gives a common coordinate value. Together these give $p\in\bigcap_jQ_j$. Every pair of points in $Q_j$ has distance at most $2M\sqrt n/2^j$, so the intersection contains only $p$. Choose $x_j\in K\cap Q_j$. Then $\|x_j-p\|\le2M\sqrt n/2^j\to0$. A closed set contains such a limit: if $p$ were in its open complement, a ball around $p$ would eventually contain $x_j$ outside $K$, a contradiction. Thus $p\in K$.

One member $O$ of the cover contains $p$, and openness gives $B(p,\varepsilon)\subseteq O$ for some $\varepsilon>0$. For sufficiently large $j$, every point of $Q_j$ is within $\varepsilon$ of $p$, so $O$ alone covers $K\cap Q_j$. This contradicts the selection of $Q_j$. Therefore every open cover has a finite subcover. The number $2^n$ is finite; replacing it by infinitely many subpieces would invalidate this argument.

Continuous images and extreme values

Theorem. If $K\subseteq\mathbb R^n$ is compact and $F:K\to\mathbb R^m$ is continuous relative to $K$, then $F(K)$ is compact. In particular, a continuous $f:K\to\mathbb R$ on nonempty compact $K$ is bounded and attains its maximum and minimum.

Proof. An open cover $(O_i)$ of $F(K)$ pulls back to relatively open sets $F^{-1}(O_i)=K\cap U_i$, with $U_i$ open in $\mathbb R^n$. These $U_i$ cover $K$, so finitely many suffice. Their corresponding $O_i$ cover $F(K)$, proving compactness. For scalar $f$, the image is nonempty, closed and bounded by the theorem already proved in dimension one. Let $a=\sup f(K)$. For each $j\ge1$, the definition of supremum supplies $y_j\in f(K)$ with $a-1/j<y_j\le a$. Then $y_j\to a$, and closedness places $a$ in the image. Thus the maximum is attained. Applying the same argument to $-f$ gives the minimum. This proves existence, not a formula for the maximizing input.

Let $d\ge1$ be finite and let $N$ be any norm on $\mathbb R^d$: it is positive except at zero, absolutely homogeneous, and satisfies the triangle inequality. Write $e_j$ for the standard coordinate vectors and $\|x\|_2=(\sum_jx_j^2)^{1/2}$. Put $C=\sum_{j=1}^dN(e_j)>0$. Expanding in the standard basis gives

The reverse triangle inequality gives $|N(x)-N(y)|\le N(x-y)\le C\|x-y\|_2$, so $N$ is Euclidean-continuous. The proved compactness theorem applies to the nonempty Euclidean unit sphere. Hence $N$ attains a minimum $m$ there; $m>0$ because every point of that sphere is nonzero. For $x\ne0$, apply this minimum to $x/\|x\|_2$. Including $x=0$, we obtain

If $(x_k)$ is $N$-Cauchy, each coordinate is real-Cauchy, since $|(x_k-x_l)_j|\le\|x_k-x_l\|_2\le N(x_k-x_l)/m$. Real completeness supplies limits $a_j$. The finite sum of the squared coordinate differences tends to zero, so $\|x_k-a\|_2\to0$ and then $N(x_k-a)\le C\|x_k-a\|_2\to0$. Thus every $N$-Cauchy sequence converges in $N$. The zero-dimensional space has just the zero vector and is complete separately. No infinite-dimensional completeness or equivalence-of-norms claim is being made.

Fix $F=\mathbb R$ or $\mathbb C$ and an integer $d\ge1$. Write $|a+ib|=\sqrt{a^2+b^2}$. This is the Euclidean norm on $\mathbb R^2$, so the proved real triangle inequality applies. Direct expansion gives $|zw|^2=|z|^2|w|^2$, hence $|zw|=|z||w|$. For $A\in M_d(F)$ define

It vanishes exactly at zero; modulus homogeneity and the scalar triangle inequality give norm homogeneity and its triangle inequality. Finite matrix multiplication gives

so $\nu(AB)\le\nu(A)\nu(B)$ and $\nu(I)=1$. Also $|a_{ij}|\le\nu(A)$.

For real matrices this is a real norm on $\mathbb R^{d^2}$. For complex matrices the map listing the real and imaginary parts of all entries identifies the space, as a real vector space, with $\mathbb R^{2d^2}$. The same $\nu$ is a real norm on those coordinates, so the proved finite-real-norm completeness theorem applies. This is an explicit realification argument, not a change of that theorem's scalar field. In either case every $\nu$-Cauchy sequence of matrices has a matrix limit in $\nu$. Matrix products preserve limits: for convergent $A_n,B_n$, boundedness of $\nu(A_n)$ and

prove the claim. The zero-dimensional matrix space is handled separately by its unique map; no extremum of an empty unit sphere is used.

Absolute matrix series and their ordered Cauchy product

If $\sum_{n\ge0}\nu(A_n)<\infty$, finite tails of $\sum A_n$ have norm at most the corresponding scalar tails. Completeness therefore gives a matrix sum. Passing a finite norm inequality to a limit is legitimate because $|\nu(X)-\nu(Y)|\le\nu(X-Y)$. Suppose also $\sum\nu(B_n)<\infty$, and let the two sums be $A,B$, with absolute sums $\alpha,\beta$. Define the finite coefficients

For each $N$, $\sum_{n=0}^N\nu(C_n)\le\sum_{k+j\le N}\nu(A_k)\nu(B_j)\le\alpha\beta$. Thus the nonnegative scalar partial sums are bounded and converge; $\sum C_n$ exists absolutely. To identify its value, put $A^{[m]}=\sum_{k=0}^m A_k$, and similarly $B^{[m]}$. When $N\ge2m$, the triangle $k+j\le N$ contains the square $k,j\le m$. Every extra pair has $k>m$ or $j>m$, so

The right side tends to zero, and $A^{[m]}B^{[m]}\to AB$. This proves $\sum C_n=AB$ without changing the order of any matrix factor or exchanging unproved double limits. Only ordinary completeness of real scalar sequences and finite sums were used.

The matrix exponential: construction, product and actual derivative

For $B\in M_d(F)$ define $\operatorname{Exp}(B)=\sum_{n=0}^{\infty}B^n/n!$, where $B^0=I$. If $B=0$, the series equals $I$ directly. Otherwise the positive scalar majorant $\sum\nu(B)^n/n!$ converges: once $n+1\ge2\nu(B)$ the ratio is at most $1/2$, so every remaining tail is bounded by a geometric tail. This proves absolute existence by the previous result.

If $B,C$ commute, the preceding Cauchy-product theorem and the finite binomial identity give

The finite binomial identity follows by induction, moving $B$ past $C$ only using $BC=CB$; it is not asserted for noncommuting matrices. In particular, for real $s,t$, $E_A(t)=\operatorname{Exp}(tA)$ satisfies $E_A(t+s)=E_A(t)E_A(s)$, $E_A(0)=I$ and $E_A(-t)E_A(t)=I$.

Here is a direct differentiation proof. Put $q=\nu(A)$. If $q=0$, $E_A=I$ and all derivative claims follow directly. Otherwise, for nonzero real $h$ with $|h|q\le1$, using $n!\ge2^{n-1}$ for $n\ge2$ gives

The product law now yields

Thus $E_A'(t)=E_A(t)A=AE_A(t)$ for real $t$, with derivative understood entrywise over $\mathbb R$ for complex entries. Commutation with $A$ follows from the finite partial sums and continuity of multiplication. This argument never assumes an integration theorem for a possibly unproved limit.

Termwise differentiation is also actually justified. On each fixed $|t|\le M$, the polynomial partial sums have tails bounded by $\sum (Mq)^n/n!$, uniformly in $t$. Their derivative tails are bounded by $q\sum_{n\ge1}(Mq)^{n-1}/(n-1)!$. Both converge by the same geometric-ratio argument, including $M=0$ separately. The derivative partial sums consequently converge uniformly to $A E_A(t)$, exactly the derivative independently proved above. Scalar complex exponentials mean the same construction in dimension one: $e^z=\sum z^n/n!$. No real-only exponential profile has been relabelled as complex.

The real zero-derivative lemma needed for uniqueness

The already selected finite-Euclidean compactness/extreme-value proof supplies extrema for a continuous real function on $[a,b]$. At an interior differentiable extremum the left and right difference quotients force its derivative to be zero. A continuous $g$ with equal endpoint values either is constant or has an interior maximum or minimum; hence, if differentiable inside, it has a zero derivative there (Rolle's argument). For a real differentiable $f$, subtract its endpoint secant line and apply this argument: $f(b)-f(a)=f'(c)(b-a)$ for some $c\in(a,b)$. Thus a real function with derivative zero on an interval is constant. Applying this separately to real and imaginary parts proves the same constant conclusion for complex-valued functions of a real parameter. There is no complex mean-value identity being assumed.

Nilpotent factors, similarity and the initial-value solution

For an integer $r\ge1$, $N^r=0$ and $\lambda\in F$, the matrices $\lambda tI$ and $tN$ commute. The proved product law and the terminating nilpotent series give, for real $t$,

This includes $\lambda=0$ with the ordinary constant term one. If $P$ is invertible and $A=PJP^{-1}$, finite powers satisfy $A^n=PJ^nP^{-1}$; the absolute limit and product continuity give $E_A(t)=PE_J(t)P^{-1}$. Hence $x(t)=E_A(t)x_0$ is differentiable, solves $x'=Ax$, and has initial value $x_0$ at zero. For completeness it is the only differentiable solution on any open real interval containing zero: if $y'=Ay$, differentiate $E_A(-t)y(t)$. Its derivative is zero, since $A$ commutes with $E_A(-t)$, and the preceding real zero-derivative lemma makes it constant equal to $y(0)$. Multiplying by $E_A(t)$ yields $y(t)=E_A(t)y(0)$. This verifies the displayed Jordan exponential formula and its full initial-value meaning over both coefficient fields, without borrowing an unadmitted matrix-ODE theorem.