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Let $N \in M_n(\mathbb{F})$ be nilpotent.
Prove that for any $1 \leq k \in \mathbb{N}$ a matrix $B \in M_n(\mathbb{F})$ exists such that $B^k=I+N$

I have no idea how to het started here. We've just covered Rational form and JCF. Any ideas? Suggestions?

JimmyK4542
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user114138
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  • I don't know the field (I presume) $\Bbb F$, but can it have characteristic $p\neq0$? – Marc van Leeuwen Dec 26 '14 at 09:38
  • It seems that it suffices to prove the claim when $N$ is a Jordan cell. – Alex Ravsky Dec 26 '14 at 09:40
  • I don't know, I presume they want us to prove it for any $\mathbb{F}$ – user114138 Dec 26 '14 at 09:40
  • can you elaborate? @AlexRavsky – user114138 Dec 26 '14 at 09:42
  • ^To elaborate on Alex's comment: If $N$ has Jordan form $N = VJV^{-1}$, then we can first find a matrix $C$ such that $C^k = I+J$, and then set $B = VCV^{-1}$. Furthermore, if the matrix $J$ in block form is $J = \text{diag}(J_1,J_2,\ldots,J_p)$, then we could just solve $C_i^k = I+J_i$ for each $i$ and set $C = \text{diag}(C_1,C_2,\ldots,C_p)$. Hence, it is sufficient to know how to solve $C_i^k = I+J$ where $J$ is a single jordan block with eigenvalue $0$. – JimmyK4542 Dec 26 '14 at 09:48
  • @JimmyK4542 Thanks, this I meant. But I am not sure, can any nilpotent matrix over any field $\Bbb F$ be transformed to Jordan form, or $\Bbb F$ should be, for instance, algebraically closed. – Alex Ravsky Dec 26 '14 at 10:00
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    @AlexRavsky: Every nilpotent matrix is similar over its defining field to a nilpotent JNF $J$. However solving the equation $C^k=I+J$ is not always possible. In fact the JNF does not seem particularly helpful in solving this equation. – Marc van Leeuwen Dec 26 '14 at 10:15
  • @MarcvanLeeuwen Thanks. – Alex Ravsky Dec 26 '14 at 10:43

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Assuming that you are working over a field of characteristic$~0$, you can use the binomial formula for exponent $\frac1k$. That is, as a series in $N$, which terminates because $N$ is nilpotent, one has that $$ B=(I+N)^{\frac1k}\buildrel{\rm def}\over=\sum_{i\in\Bbb N}\binom{1/k}iN^i $$ is a solution to $B^k=I+N$. Moreover this is the unique solution that is a polynomial in $N$.

Here one takes by definition $\binom xi=\frac{x(x_1)\ldots,(x-i+1)}{i!}$, which is used above for $x=\frac1k$. Given the size of the nilpotent matrix$~N$, the sum over $i$ is equivalent to the finite sum $\sum_{i=0}^{n-1}$.

Without the hypothesis of characteristic$~0$ a solution does not always exist. For a simple case, take $k=p=\operatorname{char}(\Bbb F)$ and write $B=I+A$, then $B^k=(I+A)^p=I+A^p$, so one has to solve $A^p=N$, but this has no solution if $N$ has a single Jordan block of size $n>1$. (It will have a solution only for very few nilpotent matrices: $A$ must be nilpotent, and the size of each Jordan blocks of $A$ is split into $p$ almost equal-size pieces giving sizes of Jordan blocks of $N$; this is a severe constraint on the Jordan type of $N$.)

Marc van Leeuwen
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