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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

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  • $\begingroup$ Hint: the characteristic polynomial is quadratic. How can we simplify higher powers of the matrix exponential with this? $\endgroup$
    – user600016
    Mar 13 at 17:39
  • $\begingroup$ @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? $\endgroup$
    – Manjoy Das
    Mar 13 at 17:49
  • $\begingroup$ By definition, $e^A=I+A+\frac{A^2}{2!}+\frac{A^3}{3!}+\dots$. $\endgroup$
    – Berci
    Mar 13 at 19:45
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There are two parts to the question:

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.

You can go from here.

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  • $\begingroup$ $e^A$ 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? $\endgroup$
    – Manjoy Das
    Mar 14 at 18:21

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