I have a 2x2 matrix A with rows (1 0) and (1 1). How can I find their matrix exponential, e^A ?

I understand I need to plug it into the Taylor series but I'm lost at how to solve the series here.

up vote 1 down vote accepted

Note. I shall do this with the series because that's what you asked for. There are easier methods which I expect you will learn later.

We have $A=I+N$ where $$N=\pmatrix{0&0\cr1&0\cr}\ .$$ It is easy to calculate that $N^2=0$. Also, $I$ and $N$ commute, so we can use the binomial theorem: $$A^k=(I+N)^k=I^k+kI^{k-1}N+\binom k2 I^{k-2}N^2+\cdots=I+kN\ .$$ So by using the series, $$\eqalign{e^A &=I+\frac1{1!}A+\frac1{2!}A^2+\frac1{3!}A^3+\cdots\cr &=I+\frac1{1!}(I+N)+\frac1{2!}(I+2N)+\frac1{3!}(I+3N)+\cdots\cr &=xI+yN\ ,\cr}$$ where $$x=1+\frac1{1!}+\frac1{2!}+\frac1{3!}+\cdots=e$$ and $$y=\frac1{1!}+\frac2{2!}+\frac3{3!}+\cdots=1+\frac1{1!}+\frac1{2!}+\cdots=e \ .$$ Hence, $$e^A=eI+eN=eA\ .$$

  • Wow thank you very much. That really made things simpler. – user191141 Nov 10 '14 at 6:51

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