Find matrix square roots

linear algebramatricesmatrix equations

I recently came across this exercise where you are supposed to find the set of all square roots of the following matrices:

  1. $\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$

  2. $\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}$

  3. $\begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$

First, I'll explain the easy way I did it: the first is pretty obvious that there are no solutions (and a well known example). For the second one, its Jordan submatrix for the eigenvalue $0$ is just the $2\times 2$ block above which can't be factored on its own.

I am unsure about the third one, since I don't know how to prove that there aren't any other solutions than the two obvious ones which are $\begin{pmatrix} 0 & ±1 & 0 \\ 0 & 0 & ±1 \\ 0 & 0 & 0 \end{pmatrix}$.

So, my $3$ questions are: are there any other square roots for the third example, if not why? how to do this just using simpler linear algebra (this was a beginning of the course exercise), without just brute forcing the system of equations? Are there any good results about the number of solutions of the equation $X^2=A$ for singular matrices (or even non-diagonalisable)?

Thanks

Best Answer

A $3 \times 3$ matrix commutes with $A = \pmatrix{0 & 0 & 1\cr 0 & 0 & 0\cr 0 & 0 & 0\cr}$ iff it is upper triangular with equal $(1,1)$ and $(3,3)$ entries. If it's a square root of $A$ the eigenvalues (which are the diagonal entries) must be $0$. And then the requirement for the square of $B = \pmatrix{0 & b_{12} & b_{13}\cr 0 & 0 & b_{23}\cr 0 & 0 & 0\cr}$ to be $A$ is $b_{12} b_{23} = 1$.

EDIT: As for $X^3 = A$, that can have infinitely many solutions as well, if the matrices are at least $4 \times 4$. Thus $$\pmatrix{0 & x_{12} & x_{13} & x_{14}\cr 0 & 0 & x_{23} & x_{24}\cr 0 & 0 & 0 & x_{34}\cr 0 & 0 & 0 & 0\cr}^3 = \pmatrix{0 & 0 & 0 & 1\cr 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0\cr}$$ iff $x_{12} x_{23} x_{34} = 1$.

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