How do I find the square root of $I+N$? Here is the question: 
Define $N \in L (F^5)$ by $N(x_1,x_2,x_3,x_4,x_5)=(2x_2,3x_3,-x_4,4x_5,0)$ Then I need to find a square root of $I+N$.
I know that first I need to find:
$N^2 (x_1,x_2,x_3,x_4,x_5)=(6x_3,-3x_4,-4x_5,0,0)$
$N^3 (x_1,x_2,x_3,x_4,x_5)=(-6x_4,-12x_5,0,0,0)$
$N^4 (x_1,x_2,x_3,x_4,x_5)=(-24x_5,0,0,0,0,)$
$N^5 (x_1,x_2,x_3,x_4,x_5)=(0,0,0,0,0)$
But how should I now the square root of $I+N$ is $I+ N/2 - N^2/8 + N^3/16 - 5N^4/128$ 
I really do not know how to figure out the last step. How do I plug this into Taylor series? 
So the denominators are basically $1/2, -1/8, 1/16$ everytime since we set $2a_1=1$ in Taylor series?
 A: For real numbers, where we can talk about convergence, we have the (possibly well-known) Taylor series
$$\tag1\sqrt{1+x}= 1 + \frac12x - \frac18x^2 + \frac1{16}x^3 - \frac5{128}x^4 + \frac7{256}x^5 - \frac{21}{1024}x^6 + \ldots $$
for all sufficiently small $x$ (namely those with $x^n\to 0$).
Of course, this does not make sense here, at least not directly. However, we have "convergence" of $N^n\to 0$ here as well, even if we do not have any metric available: $N^5=0$ and so $N^n=0$ for $n\ge 5$.
If we chop the right hand side at the $x^5$ term, then squaring the chopped series, we will not obtain $1+x$, but rather $1+x+a_5x^5+a_6x^6+\ldots$. However, if we do the same with $N$ instead of $x$ (and of course $I$ instead of $1$), the higher powers are all zero, hence can be dropped, and the square of the chopped series is really $I+N$.
Remark: Of course one thing must be noted: The above does not work if the characteristic of $F$ is $2$, for in that case it does not make sense to speak of $\frac 12$, $\frac 18$, etc.
And indeed, in characteristic $2$, $(I+N)(x_1,x_2,x_3,x_4,x_5)=(x_1,x_2+x_3,x_3+x_4,x_4,x_5)$ looks like not having a square root (or does it?).
