$X^2 = L$, with $L$ lower triangular, but $X$ is not lower triangular. Is it possible?
I know that a lower triangular matrix $L$ (not a diagonal matrix for this question),
$$L_{nm} \cases{=0 & for all $m > n$ \\ \ne 0 & for some $ m<n$} $$
when squared is lower triangular. But is the square root, when it exists, always lower triangular? I have found some examples that give a diagonal matrix:
$$\pmatrix{1 & 0 & 1 \\ 0 & e & 0 \\ c & 0 & -1}\pmatrix{1 & 0 & 1 \\ 0 & e & 0 \\ c & 0 & -1}=\pmatrix{c+1 & 0 & 0 \\ 0 & e^2 & 0 \\ 0 & 0 & c+1}=L$$
But I am wondering about the possibility of the square being strictly triangular, not diagonal.
I believe the answer is yes, that if the square root of a strictly lower triangular exists, then that is also lower triangular. I am looking for a good argument as to why, or for any counter examples.
EDIT:
Also, all $ 2 \times 2$ that gives the upper right zero in the square implies also the lower left is zero.
$$\pmatrix{a & c \\ d & b\\}\pmatrix{a & c \\ d & b\\}=\pmatrix{a^2 + cd & ac + cb \\ ad + bd & cd + b^2\\}$$
$$(ac+cb= 0) \Rightarrow (a = -b) \Rightarrow (ad+bd=0)$$
So a counter example will necessarily be higher dimension, but I am thinking that the same logic will still apply somehow, along the lines of $2 \times 2$ sub-matrices or something.
Best Answer
$$ \left( \begin{matrix} 1 & 1 & 1 \\ 4 & 1 & 2 \\ 1 & -2 & -3 \end{matrix} \right)^2 = \left( \begin{matrix} 6 & 0 & 0 \\ 10 & 1 & 0 \\ -10 & 5 & 6 \end{matrix} \right)$$
This example was found more or less by