[proofplan] We work in the quotient algebra $\mathbb{R}[z]/(p_A)$, where $p_A$ is the characteristic polynomial of $A$. Since $p_A$ is monic of degree $n$, every element of this quotient has a unique representative of degree at most $n-1$, so the quotient exponential of the class of $tz$ has unique coordinate functions $\alpha_0(t),\dots,\alpha_{n-1}(t)$. The [Cayley-Hamilton theorem](/theorems/923) makes evaluation at $A$ descend from $\mathbb{R}[z]$ to the quotient, and applying this evaluation map to the quotient exponential gives the ordinary matrix exponential. [/proofplan]

[step:Build the quotient algebra where all powers reduce to degree less than $n$]Let $I_n \in \mathbb{R}^{n \times n}$ denote the identity matrix. Let $p_A \in \mathbb{R}[z]$ denote the characteristic polynomial of $A$, defined by \begin{align*} p_A(z) = \det(z I_n - A). \end{align*} Then $p_A$ is monic of degree $n$. Define the real quotient algebra \begin{align*} Q_A := \mathbb{R}[z]/(p_A), \end{align*} and let \begin{align*} \pi_A: \mathbb{R}[z] \to Q_A \end{align*} be the quotient map. Define \begin{align*} \xi := \pi_A(z) \in Q_A. \end{align*} Because $p_A$ is monic of degree $n$, polynomial division gives the following uniqueness statement: for every $f \in \mathbb{R}[z]$, there are unique polynomials $q_f,r_f \in \mathbb{R}[z]$ such that $\deg r_f < n$ and \begin{align*} f = q_f p_A + r_f. \end{align*} Therefore every class in $Q_A$ has a unique representative of the form \begin{align*} c_0 + c_1 z + \cdots + c_{n-1} z^{n-1}, \end{align*} with $c_0,\dots,c_{n-1} \in \mathbb{R}$. Hence \begin{align*} \pi_A(1), \pi_A(z), \dots, \pi_A(z^{n-1}) \end{align*} is a basis of the finite-dimensional real [vector space](/page/Vector%20Space) $Q_A$.[/step]

[guided]The goal is to make the phrase “reduce powers of $A$ using the characteristic polynomial” into a precise algebraic construction. We first perform the reduction before substituting $A$. Let $I_n \in \mathbb{R}^{n \times n}$ denote the identity matrix. Let $p_A \in \mathbb{R}[z]$ be the characteristic polynomial of $A$: \begin{align*} p_A(z) = \det(z I_n - A). \end{align*} This polynomial is monic of degree $n$. We form the quotient algebra \begin{align*} Q_A := \mathbb{R}[z]/(p_A), \end{align*} and write \begin{align*} \pi_A: \mathbb{R}[z] \to Q_A \end{align*} for the quotient map. The element that plays the role of the variable inside this quotient is \begin{align*} \xi := \pi_A(z). \end{align*} Why use this quotient? In $Q_A$, the polynomial $p_A$ is declared to be zero. Since $p_A$ has degree $n$, this forces every sufficiently high power of $\xi$ to be expressible as a linear combination of $1,\xi,\dots,\xi^{n-1}$. We now justify that this reduction is unique. Since $p_A$ is monic of degree $n$, the Euclidean division theorem in $\mathbb{R}[z]$ says that for every polynomial $f \in \mathbb{R}[z]$, there exist unique polynomials $q_f,r_f \in \mathbb{R}[z]$ with $\deg r_f < n$ such that \begin{align*} f = q_f p_A + r_f. \end{align*} Passing to the quotient gives \begin{align*} \pi_A(f) = \pi_A(r_f), \end{align*} because $\pi_A(q_f p_A)=0$. Thus every class has a representative of degree less than $n$. Uniqueness also matters. If two polynomials $r,s \in \mathbb{R}[z]$ both have degree less than $n$ and represent the same class in $Q_A$, then $r-s$ is divisible by $p_A$. But either $r-s=0$, or $\deg(r-s)<n$, while every nonzero multiple of the monic degree-$n$ polynomial $p_A$ has degree at least $n$. Therefore $r=s$. Hence \begin{align*} \pi_A(1), \pi_A(z), \dots, \pi_A(z^{n-1}) \end{align*} is a basis of $Q_A$ as a real vector space.[/guided]

[step:Define the scalar coefficient functions from the quotient exponential]For each $t \in \mathbb{R}$, define the quotient-algebra exponential \begin{align*} \operatorname{Exp}_{Q_A}(t\xi) := \sum_{m=0}^{\infty} \frac{t^m \xi^m}{m!} \in Q_A. \end{align*} Equip the finite-dimensional real vector space $Q_A$ with the norm $\|\cdot\|_{Q_A}$ obtained from the basis $\pi_A(1),\pi_A(z),\dots,\pi_A(z^{n-1})$ by identifying $Q_A$ with $\mathbb{R}^n$ and using the Euclidean norm. Let $M_\xi: Q_A \to Q_A$ be the [linear map](/page/Linear%20Map) $u \mapsto \xi u$. Since $M_\xi$ is a linear map on a finite-dimensional normed space, its operator norm $\|M_\xi\|_{\mathrm{op}}$ is finite. Therefore \begin{align*} \|\xi^m\|_{Q_A} \leq \|M_\xi\|_{\mathrm{op}}^m \|1\|_{Q_A} \end{align*} for every integer $m \geq 0$, so the series defining $\operatorname{Exp}_{Q_A}(t\xi)$ is dominated by the convergent real exponential series \begin{align*} \|1\|_{Q_A}\sum_{m=0}^{\infty}\frac{(|t|\|M_\xi\|_{\mathrm{op}})^m}{m!}. \end{align*} Thus the quotient exponential is well-defined in $Q_A$. Since $\pi_A(1),\pi_A(z),\dots,\pi_A(z^{n-1})$ is a basis of $Q_A$, for each $t \in \mathbb{R}$ there exist unique [real numbers](/page/Real%20Numbers) $\alpha_0(t),\dots,\alpha_{n-1}(t)$ such that \begin{align*} \operatorname{Exp}_{Q_A}(t\xi) = \sum_{k=0}^{n-1} \alpha_k(t)\xi^k. \end{align*} This defines functions \begin{align*} \alpha_k: \mathbb{R} \to \mathbb{R} \end{align*} for $k=0,\dots,n-1$.[/step]

[guided]Now that every element of $Q_A$ has unique coordinates in the basis $\pi_A(1),\pi_A(z),\dots,\pi_A(z^{n-1})$, we define the coefficients by taking the exponential inside this finite-dimensional algebra. For $t \in \mathbb{R}$, set \begin{align*} \operatorname{Exp}_{Q_A}(t\xi) := \sum_{m=0}^{\infty} \frac{t^m \xi^m}{m!} \in Q_A. \end{align*} We must justify that this infinite series converges in $Q_A$, because $Q_A$ is not merely a formal quotient at this point; it is being used as a finite-dimensional [normed vector space](/page/Normed%20Vector%20Space). Define $\|\cdot\|_{Q_A}$ by identifying $Q_A$ with $\mathbb{R}^n$ through the ordered basis $\pi_A(1),\pi_A(z),\dots,\pi_A(z^{n-1})$ and using the Euclidean norm on $\mathbb{R}^n$. Let $M_\xi: Q_A \to Q_A$ be the multiplication map $u \mapsto \xi u$. This is a linear map on the finite-dimensional normed real vector space $Q_A$, so its operator norm $\|M_\xi\|_{\mathrm{op}}$ is finite. Since $\xi^m = M_\xi^m(1)$, repeated use of the operator norm gives \begin{align*} \|\xi^m\|_{Q_A} \leq \|M_\xi\|_{\mathrm{op}}^m \|1\|_{Q_A} \end{align*} for every integer $m \geq 0$. Therefore \begin{align*} \left\|\frac{t^m\xi^m}{m!}\right\|_{Q_A} \leq \|1\|_{Q_A}\frac{(|t|\|M_\xi\|_{\mathrm{op}})^m}{m!}. \end{align*} The scalar majorant series is the ordinary exponential series, hence it converges. By comparison in the finite-dimensional normed space $Q_A$, the series defining $\operatorname{Exp}_{Q_A}(t\xi)$ converges. The reason for introducing this exponential is that it has coordinates in the reduced-power basis. Since $\pi_A(1),\pi_A(z),\dots,\pi_A(z^{n-1})$ is a basis of $Q_A$, each element of $Q_A$ has a unique coordinate expansion. Applying this to $\operatorname{Exp}_{Q_A}(t\xi)$ gives unique real numbers $\alpha_0(t),\dots,\alpha_{n-1}(t)$ such that \begin{align*} \operatorname{Exp}_{Q_A}(t\xi) = \sum_{k=0}^{n-1} \alpha_k(t)\xi^k. \end{align*} Thus, for each $k=0,\dots,n-1$, we have defined a scalar function \begin{align*} \alpha_k: \mathbb{R} \to \mathbb{R}. \end{align*}[/guided]

[step:Descend polynomial evaluation at $A$ to the quotient]Define the polynomial evaluation map \begin{align*} \operatorname{ev}_A: \mathbb{R}[z] \to \mathbb{R}^{n \times n} \end{align*} by \begin{align*} \operatorname{ev}_A\left(\sum_{j=0}^{m} c_j z^j\right) = \sum_{j=0}^{m} c_j A^j. \end{align*} By the [Cayley-Hamilton Theorem](/theorems/865), applied to the matrix $A \in \mathbb{R}^{n \times n}$ with characteristic polynomial $p_A$, we have $p_A(A)=0$. Hence, if $f-g \in (p_A)$, then $f-g = h p_A$ for some $h \in \mathbb{R}[z]$, and multiplicativity of polynomial evaluation gives \begin{align*} \operatorname{ev}_A(f)-\operatorname{ev}_A(g) = \operatorname{ev}_A(h)\operatorname{ev}_A(p_A) = \operatorname{ev}_A(h)p_A(A) = 0. \end{align*} Therefore evaluation at $A$ is constant on congruence classes modulo $(p_A)$, so it induces a well-defined real algebra homomorphism \begin{align*} \widetilde{\operatorname{ev}}_A: Q_A \to \mathbb{R}^{n \times n} \end{align*} given by \begin{align*} \widetilde{\operatorname{ev}}_A(\pi_A(f)) = f(A) \end{align*} for every $f \in \mathbb{R}[z]$. In particular, \begin{align*} \widetilde{\operatorname{ev}}_A(\xi^m)=A^m \end{align*} for every integer $m \ge 0$.[/step]

[guided]We now explain why substituting $A$ is compatible with the quotient. Define the polynomial evaluation map \begin{align*} \operatorname{ev}_A: \mathbb{R}[z] \to \mathbb{R}^{n \times n} \end{align*} by \begin{align*} \operatorname{ev}_A\left(\sum_{j=0}^{m} c_j z^j\right) = \sum_{j=0}^{m} c_j A^j. \end{align*} This is a real algebra homomorphism because evaluation respects addition and multiplication of polynomials. To descend this map to $Q_A=\mathbb{R}[z]/(p_A)$, we must prove that two representatives of the same quotient class have the same value at $A$. Suppose $f,g \in \mathbb{R}[z]$ represent the same class. Then $f-g \in (p_A)$, so there exists $h \in \mathbb{R}[z]$ such that \begin{align*} f-g = h p_A. \end{align*} The [Cayley-Hamilton Theorem](/theorems/865), applied to the matrix $A$ and its characteristic polynomial $p_A$, gives \begin{align*} p_A(A)=0. \end{align*} Using multiplicativity of polynomial evaluation, we get \begin{align*} \operatorname{ev}_A(f)-\operatorname{ev}_A(g) = \operatorname{ev}_A(h)\operatorname{ev}_A(p_A) = \operatorname{ev}_A(h)p_A(A) = 0. \end{align*} Therefore evaluation at $A$ depends only on the quotient class. Hence there is a well-defined real algebra homomorphism \begin{align*} \widetilde{\operatorname{ev}}_A: Q_A \to \mathbb{R}^{n \times n} \end{align*} satisfying \begin{align*} \widetilde{\operatorname{ev}}_A(\pi_A(f)) = f(A) \end{align*} for every $f \in \mathbb{R}[z]$. Since $\xi=\pi_A(z)$, this gives \begin{align*} \widetilde{\operatorname{ev}}_A(\xi^m)=A^m \end{align*} for every integer $m \geq 0$.[/guided]

[step:Evaluate the quotient exponential to obtain the matrix exponential]Fix $t \in \mathbb{R}$. Since $\widetilde{\operatorname{ev}}_A: Q_A \to \mathbb{R}^{n \times n}$ is a linear map between finite-dimensional normed real vector spaces, it is continuous and may be applied term-by-term to convergent series. Therefore \begin{align*} \widetilde{\operatorname{ev}}_A\left(\operatorname{Exp}_{Q_A}(t\xi)\right) = \widetilde{\operatorname{ev}}_A\left(\sum_{m=0}^{\infty} \frac{t^m\xi^m}{m!}\right) = \sum_{m=0}^{\infty} \frac{t^m\widetilde{\operatorname{ev}}_A(\xi^m)}{m!}. \end{align*} Using $\widetilde{\operatorname{ev}}_A(\xi^m)=A^m$, this becomes \begin{align*} \widetilde{\operatorname{ev}}_A\left(\operatorname{Exp}_{Q_A}(t\xi)\right) = \sum_{m=0}^{\infty} \frac{t^m A^m}{m!} = e^{tA}, \end{align*} where the final equality is the definition of the matrix exponential. On the other hand, applying $\widetilde{\operatorname{ev}}_A$ to the coordinate expansion of $\operatorname{Exp}_{Q_A}(t\xi)$ gives \begin{align*} \widetilde{\operatorname{ev}}_A\left(\operatorname{Exp}_{Q_A}(t\xi)\right) = \widetilde{\operatorname{ev}}_A\left(\sum_{k=0}^{n-1}\alpha_k(t)\xi^k\right) = \sum_{k=0}^{n-1}\alpha_k(t)A^k. \end{align*} Combining the two displayed identities yields \begin{align*} e^{tA} = \sum_{k=0}^{n-1}\alpha_k(t)A^k. \end{align*} Since $t \in \mathbb{R}$ was arbitrary, the identity holds for every $t \in \mathbb{R}$.[/step]

[guided]It remains to connect the quotient construction back to the matrix exponential. Fix $t \in \mathbb{R}$. The map \begin{align*} \widetilde{\operatorname{ev}}_A: Q_A \to \mathbb{R}^{n \times n} \end{align*} is linear between finite-dimensional normed real vector spaces, hence continuous. Therefore it may be applied term-by-term to the convergent series defining $\operatorname{Exp}_{Q_A}(t\xi)$. We obtain \begin{align*} \widetilde{\operatorname{ev}}_A\left(\operatorname{Exp}_{Q_A}(t\xi)\right) = \widetilde{\operatorname{ev}}_A\left(\sum_{m=0}^{\infty} \frac{t^m\xi^m}{m!}\right) = \sum_{m=0}^{\infty} \frac{t^m\widetilde{\operatorname{ev}}_A(\xi^m)}{m!}. \end{align*} The previous step proved that $\widetilde{\operatorname{ev}}_A(\xi^m)=A^m$ for every integer $m \geq 0$, so \begin{align*} \widetilde{\operatorname{ev}}_A\left(\operatorname{Exp}_{Q_A}(t\xi)\right) = \sum_{m=0}^{\infty} \frac{t^m A^m}{m!} = e^{tA}. \end{align*} The last equality is the definition of the matrix exponential. We also evaluate the coordinate expansion that defined the coefficient functions. Since \begin{align*} \operatorname{Exp}_{Q_A}(t\xi) = \sum_{k=0}^{n-1}\alpha_k(t)\xi^k, \end{align*} linearity of $\widetilde{\operatorname{ev}}_A$ gives \begin{align*} \widetilde{\operatorname{ev}}_A\left(\operatorname{Exp}_{Q_A}(t\xi)\right) = \sum_{k=0}^{n-1}\alpha_k(t)\widetilde{\operatorname{ev}}_A(\xi^k) = \sum_{k=0}^{n-1}\alpha_k(t)A^k. \end{align*} Both displayed formulas compute the same matrix $\widetilde{\operatorname{ev}}_A(\operatorname{Exp}_{Q_A}(t\xi))$. Therefore \begin{align*} e^{tA} = \sum_{k=0}^{n-1}\alpha_k(t)A^k. \end{align*} Since $t \in \mathbb{R}$ was arbitrary, the identity holds for all real $t$.[/guided]