Does an invertible real matrix have a real cubic root

Question

$A$ is an invertible real matrix, is there a real matrix $B$ such that $A=B^3$?

For the complex case, it is not difficult. Over complex numbers, every invertible matrix has a cubic root. In fact, over $\mathbb C$, since every invertible matrix has a logrithm, we can take a family of matrices $e^{t\log A}$, and taking $t=1/3$ yields a square root of $A$.

The condition "invertible " is necessary for the real case. In fact, here's a counterexample link.

How about general cases? $A$ is an invertible real matrix, is there a real matrix $B$ such that $A=B^n$, where integer $n\geq 2$?

Answer 1

Yes. It suffices to prove that $A$ is similar to a real matrix $C$ that can easily be seen to have a real cubic root $R$. So, if $P^{-1}CP=A$ and $R^3=C$, we may take $B=P^{-1}RP$.

Since all real matrices admit “real Jordan forms” (please consult Matrix Analysis by Horn and Johnson if you don't know what a real Jordan form is), it suffices to consider the case where $A$ is a real Jordan form with a single Jordan block. That is, we may assume that $$ A=J_k(\lambda)=\pmatrix{\lambda&1\\ &\ddots&\ddots\\ &&\ddots&1\\ &&&\lambda} \quad\text{or}\quad A=J_k(rQ_\theta)=\pmatrix{rQ_\theta&I_2\\ &\ddots&\ddots\\ &&\ddots&I_2\\ &&&rQ_\theta} $$ where $\lambda$ is a nonzero real number, $r>0$ and $Q_\theta\ne\pm I_2$ is the $2\times2$ rotation matrix for angle $\theta$.

If $A=J_k(\lambda)$, take $R=J_k(\lambda^{1/3})$. Then $C$ and $A$ are similar because both of them are non-derogatory matrices with the same minimal polynomial $(x-\lambda)^k$. (That $\lambda\ne0$ is essential here. If $\lambda=0$, the minimal polynomial of $C=R^3$ is no longer $(x-\lambda)^k$ unless $k=1$).

Similarly, if $A=J_k(rQ_\theta)$, take $R=J_k(r^{1/3}Q_{\theta/3})$. Note that $A$ and $R$ are similar over $\mathbb C$ to $$ J_k(re^{i\theta})\oplus J_k(re^{-i\theta}) \quad\text{and}\quad J_k(r^{1/3}e^{i\theta/3})\oplus J_k(r^{1/3}e^{-i\theta/3}) $$ respectively. Thus the minimal polynomials of $A$ and $C$ (as $r\ne0$) are both equal to $(x-re^{i\theta})^k(x-re^{-i\theta})^k=\left(x^2-2r\cos(\theta)x+r^2\right)^k$. So, the two matrices are again similar because they are non-derogatory matrices with the same minimal polynomial.

Answer 2

Yes, every real matrix that is diagonalizable over the complex numbers has a real cube root. Suppose $A$ is such a matrix. Then there is a real invertible matrix $P$ such that $P^{-1}AP =D$ where $D$ is a matrix that is 'almost' diagonal-it has real numbers $\lambda_1,...,\lambda_m$ along the main diagonal and also (if it has one or more non-real eigenvalues) some 2x2 real blocks of the form $$\begin{bmatrix}a&-b\\b&a\end{bmatrix}=\begin{bmatrix}r\cos\theta&-r\sin\theta\\r\sin\theta&r\cos\theta\end{bmatrix}$$along the main diagonal and 0's elsewhere.Let $E$ be the matrix in which each $\lambda_i$ is replaced by $\lambda_i^{1/3}$ and each 2x2 block is replaced by $$r^{1/3}\begin{bmatrix}\cos(\theta/3)&-\sin(\theta/3)\\\sin(\theta/3)&\cos(\theta/3)\end{bmatrix}$$. Then $(PEP^{-1})^3=A$. Similarly the matrix $A$ has $\nu$-th roots if $\nu$ is any odd natural number.

Answer 3

For square roots of real $A$ to exist, it is enough that all of the real eingenvalues of $A$ are $>0$.

Cubic roots, and more generally roots of odd order, exist for every real invertible $A$. The proof can be done like in the solution of @user1551. A similar approach is using polynomials that do Hermite interpolation of the function $t\mapsto\sqrt[3]{t}$ on the spectrum of $A$ ( considered with multiplicities).

I tested an approach for cubic roots using the Newton method. That is, start with some $x_0$ ( perhaps $I$ if not a problem), and use the recurrence

$$x_{n+1} = x_n - \frac{x_n^3-a}{3 x_n^2}$$

(the quotient since all of the $x_n$ commute with $a$). It seems that the convergence is quite good. This could work with other polynomial equations $P(B)= A$.

Another approach works if $A$ is not too far from $I$ , that is $A= I+\delta$, with $\delta$ small ( the spectral radius $\rho(\delta)<1$). Then we can take

$$\sqrt[n]{A} = \sqrt[n]{1+\delta}= (1+\delta)^{\frac{1}{n}} = \sum_{k\ge 0}\binom{\frac{1}{n}}{k} \delta^k$$