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Yo, I couldn't solve this exercise after thinking for a while.

For every $A \in GL_{3} (\mathbb{C})$ and $n$, there's a $B \in Mat_{3, 3}(\mathbb{C})$ such that $B^n = A$

The previous exercise was that for every nilpotent $N \in Mat_{3, 3} (\mathbb{C})$ and every $n$, $C = 1 + \frac{1}{n}N + \frac{1-n}{2n^2}N^2$ satisfies $C^n = 1 + N$, so I suppose there's a trick using this result.

I tried to play a little with the splitting of of $A$ as a nilpotent plus a semisimple, however I couldn't get anything useful.

Thanks in advance.

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If you write $A = D + N$ as semisimple + nilpotent (where $D$ and $N$ commute), then $D$ is invertible and $$A = D(I + D^{-1}N),$$ where $D^{-1}N$ is nilpotent (because $D$ and $N$ commute). Now $D$ has a $n$th root (because we are in $\mathbb{C}$ so it's diagonalizable), and so does $I + D^{-1}N$ by the previous exercise. The product of these two $n$th roots is your desired $B$ (because they commute).


More generally, for matrices of any size, you can put $B = \exp(\tfrac1n \log A)$. Here, $\exp$ is defined by the usual power series, and $\log A$ is any matrix such that $\exp(\log A) = A$. If $A$ is invertible then this exists. Indeed, with $A = D(I + D^{-1}N)$ as above, then $\log D$ exists (clearly we can take the logarithm of any invertible diagonal matrix), and for the other factor we can use the power series $$\log (I+X) = \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k} X^k$$ for $X = D^{-1}N$. The power series converges whenever the spectral radius of $X$ is $< 1$; in particular, it converges (after finitely many terms) when $X$ is nilpotent. Then $$\log A = \log D + \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k} (D^{-1}N)^k.$$ It's worth showing that the $C$ from your previous exercise is just $C = \exp(\tfrac1n \log(I+N))$.


Actually, it's even easier to derive $C$ using the binomial series $$ (I+X)^\alpha = \sum_{k=0}^\infty \binom{\alpha}{k} X^k. $$

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  • $\begingroup$ Sorry for replying too later. I was very busy. Thanks for your answer. $\endgroup$ – user40276 Nov 17 '16 at 20:31
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Given your $A$, write its Jordan form. Because here the matrices are $3\times 3$, there is not much room for things to happen:

1) If $A$ has three different eigenvalues, then $A=SJS^{-1}$ with $J$ diagonal. Now choose a diagonal matrix $X$ such thath $X_{jj}^2=J_{jj}$. Then $SXS^{-1}$ is is a square root of $A$.

2) If $A$ has a repeated eigenvalue, its Jordan form may still be diagonal, and 1) applies.

3) If $A$ has a $2\times2$ Jordan block, $J$, then $J=\lambda_1 I_2+N$ with $N$ nilpotent. Then $\lambda_1^{1/2}+\frac1{2\lambda_1^{1/2}}N$ is a square root for $J$; as $$A=\begin{bmatrix}J&0\\0&\lambda_2\end{bmatrix},$$ it has a square root $$\begin{bmatrix}\lambda_1^{1/2}+\frac1{2\lambda_1^{1/2}}N&0\\0&\lambda_2^{1/2}\end{bmatrix}.$$

4) If $A=SJS^{-1}$ with $J$ a $3\times 3$ Jordan block, then $J=\lambda I+N$; now the exercise you quoted shows that $$C=I+\frac N2-\frac{N^2}8$$ is a square root of $I+N$. Then $$\lambda^{1/2}\left(I+\frac N{2\lambda}-\frac{N^2}{8\lambda^2}\right)$$ is a square root for $A$.

The fact that $A$ is invertible is used to guarantee that all eigenvalues are nonzero. For instance, if $$A=\begin{bmatrix}0&1\\0&0\end{bmatrix},$$ then no $B$ exists such that $B^2=A$.

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  • $\begingroup$ Sorry for replying too later. I was very busy. In 3) $N$ is not nilpotent. The third element of the basis is an eingavalue of $N$. $\endgroup$ – user40276 Nov 17 '16 at 20:30
  • $\begingroup$ I was talking about a $2\times2$ nilpotent. I have edited point 3. $\endgroup$ – Martin Argerami Nov 17 '16 at 22:21
  • $\begingroup$ Got it. Thanks for clarifying. $\endgroup$ – user40276 Nov 17 '16 at 22:25
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If $A$ is diagonalizable, $A$ = $M^{-1}$.$D$.$M$, then $D$ obviously has a cube-root $D_{1/3}$, and $A$ = [$M^{-1}$.$D_{1/3}$.$M$$]^3$. Helps?

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  • $\begingroup$ And if it is not? $\endgroup$ – N. S. Oct 13 '16 at 21:59

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