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.


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 $$ (I+X)^\alpha = \sum_{k=0}^\infty \binom{\alpha}{k} X^k. $$

  • $\begingroup$ Sorry for replying too later. I was very busy. Thanks for your answer. $\endgroup$ – user40276 Nov 17 '16 at 20:31

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 $B^2=A$.

  • $\begingroup$ 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

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?

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.