Square root and Cubic root of a Matrix

Question

Need

• an example of a matrix with a square root but it does not admit any cubic roots

• an example of a matrix with a cubic root but it does not admit any square root

Answer 1

Hint: You can find suitable examples among the powers of Jordan blocks for $\lambda = 0$.

Answer 2

COMMENT.- The cube of $A=\begin{pmatrix}1&2&3\\4&5&6\\7&8&9\end{pmatrix}$ is equal to $B=\begin{pmatrix}468&576&684\\1062&1305&1548\\1656&2034&2412\end{pmatrix}$ so it is evident that $B$ has a cubic root equal to $A$.

In order to know if $B$ has a square root we must to find a matrix $X=\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}$ such that its square, i.e $X^2=\begin{pmatrix}a^2+bd+cg&ab+be+ch&ac+bf+ci\\da+ed+fg&bd+e^2+fh&dc+ef+fi\\ga+hd+ig&gb+he+ih&cg+fh+i^2\end{pmatrix}$ be equal to the matrix $B$. This means that we must solve a system of nine equations with nine unknowns to have a non-simple example with $0$ and $1$ convenient to facilitate the task that is guessed laborious.

But above all this should serve to take into account two important things: that it is necessary to indicate to which set the terms of the matrix belong and mainly note that not all matrices have cubic or square roots for the simple reason that the system that we must solving in general will not be compatible.In the language of functional analysis it is said that a function $f$ "operates" in an algebra $A$ if $f(a)$ makes sense for all elements $a$ of $A$. Thus, and using this language we could say that neither the square root function nor the cubic root function "operate" on the algebra of the square matrices of order $n$ (instead the exponential function $f (x) = e^x$ does operate in this algebra).