Computing $e^{At}$ in terms of matrix

Question

Suppose we have a matrix of the form

$A = I + N$ with $N^2 = 0$.

Compute $e^{At}$ in terms of $N$.

My initial attempt to solve this problem was

$e^{At} = I + At + \frac{At^2}{2!} + \frac{At^3}{3!} + \cdots$

$\;\;\;\;\; = I + A(1 + t + \frac{t^2}{2!} + \frac{t^3}{3!} + \cdots) - A$

$\;\;\;\;\; = I - A + Ae^t$

$\;\;\;\;\; = -N + (I + N)e^t$

$\;\;\;\;\; = Ie^t + N(e^t - 1)$

However, the solution that I was provided states that

by induction,

$A^k = I + (k-1)N$

$e^{At} = e^tI + te^tN$

which confuses me because I thought

$A^k = I + kN$

I am not sure what went wrong with my approach.

Could someone help me out?

Thank you for reading.

Answer 1

You missed the powers of $A$ in the first line. It should be \begin{eqnarray*} e^{At} = I + At + \frac{A^{\color{red}{2}}t^2}{2!} + \frac{A^{\color{red}{3}}t^3}{3!} + \cdots. \end{eqnarray*} With $A^k=I+kN$ should be quite easy from here ?

Answer 2

If $A$ and $B$ commute, then $$ \begin{align} e^{A+B} &=\sum_{n=0}^\infty\frac1{n!}(A+B)^n\\ &=\sum_{n=0}^\infty\frac1{n!}\sum_{k=0}^n\binom{n}{k}A^{n-k}B^k\\ &=\sum_{k=0}^\infty\sum_{n=k}^\infty\frac1{(n-k)!}A^{n-k}\frac1{k!}B^k\\ &=\sum_{k=0}^\infty\sum_{n=0}^\infty\frac1{n!}A^n\frac1{k!}B^k\\[3pt] &=e^Ae^B \end{align} $$ Therefore, since $I$ commutes with every matrix, $$ \begin{align} e^{t(I+N)} &=e^{tI}e^{tN}\\ &=e^tI(I+tN)\\ &=e^t(I+tN) \end{align} $$ Or we can $$ \begin{align} e^{t(I+N)} &=\sum_{k=0}^\infty\frac{t^k}{k!}(I+N)^k\\ &=\sum_{k=0}^\infty\frac{t^k}{k!}(I+kN)\\ &=\sum_{k=0}^\infty\frac{t^k}{k!}I+\sum_{k=1}^\infty\frac{t^k}{k!}kN\\ &=\sum_{k=0}^\infty\frac{t^k}{k!}I+\sum_{k=1}^\infty\frac{t^k}{(k-1)!}N\\ &=\sum_{k=0}^\infty\frac{t^k}{k!}I+\sum_{k=0}^\infty\frac{t^{k+1}}{k!}N\\[6pt] &=e^tI+te^tN \end{align} $$