# Representing matrix exponential in terms of matrix powers via Cayley-Hamilton

Using the Cayley-Hamilton theorem, show that for a $$2 \times 2$$ matrix $$A$$, $$e^A = c_1 A + c_0 I$$ where $$c_1$$ and $$c_2$$ are constants.

This problem came to me absolutely from nowhere and I have no clue how to solve it. Basically the theorem tells that every square matrix satisfies its own characteristic equation. By this theorem, we can find the exponent of a square matrix and its inverse. But does it really give a clue how to find anything like $$e^A$$? Help me to solve this

• Hint: the characteristic polynomial is quadratic. How can we simplify higher powers of the matrix exponential with this? Mar 13 at 17:39
• @user600016 If $P_1\lambda^2+P_2\lambda+P_3=0$ be the ch. eqn, then I can replace $\lambda$ by $A$. But after that how do I bring $e$ into my calculation? Mar 13 at 17:49
• By definition, $e^A=I+A+\frac{A^2}{2!}+\frac{A^3}{3!}+\dots$. Mar 13 at 19:45

Lemma 1 For any polynomial $$P,$$ $$P(A) = c_1(P) A + c_2(P)I.$$
Proof sketch The right hand side is equal to the remainder when $$P(x)$$ is divided by $$\chi(x),$$ where $$\chi$$ is the characteristic polynomial of $$A.$$
Lemma 2 If $$P_n$$ is the $$n$$-th Taylor polynomial of $$\exp,$$ the coefficients $$c_{1, 2}(P_n)$$ converge as $$n$$ goes to infinity.
• $e^A$ has coefficients $1, \dfrac 1 {2!}, \dfrac 1 {3!}$ etc and that of $\chi(x)$ will be $P_1, P_2, P_3$ where these are all constants. So on division, how the coefficients of the remainder will be functions of $P$? Also what will be $P(A)$ in this case? Mar 14 at 18:21