Assume $\bf A\in \mathbb R^{n\times n}$
If I know ${\bf A}^k$ and that it is symmetric, how can I systematically find the $\bf A$ which are not?
Own work One approach I have considered is to assume a Jordan form:
$${\bf A}^k = ({\bf SJS}^{-1})^k$$
Now complications will arise if there are real negative eigenvalues of multiplicity 1 in ${\bf A}^k$. So let us for simplicity assume ${\bf A}^k$ has any such eigenvalue multiplicity > 1.
In the first case we can reduce the problem to finding block-matrix k:th roots on the Jordan blocks and constructing ${\bf SMJS}^{-1}$
Where the $M$ matrix is block diagonal with these k:th roots ordered to match the block-structure of $S$.
As for a minimal example $k = 2, {\bf J} = \begin{bmatrix} 0&1\\0&0\end{bmatrix}$ we can find all possibilities for the $2\times 2$ zero matrices by varying $\bf S$ (?)
Another example $$n=3, k=16, {\bf A}^{16} = \left[\begin{array}{lll}2&1&1\\1&2&1\\1&1&2\end{array}\right]$$
$$J = \left[\begin{array}{lll}1&0&0\\0&1&0\\0&0&4\end{array}\right], S = \left[\begin{array}{lll}0.0398278888854&0.8155246200658&0.5773502691896\\-0.7261789828314&-0.3732703464790&0.5773502691896\\0.6863510939459&-0.4422542735868&0.5773502691896\end{array}\right]$$ Then devise $$M = \begin{bmatrix}\cos(2\pi/16)&\sin(2\pi/16)&0\\ -\sin(2\pi/16)&\cos(2\pi/16)&0\\ 0&0&1\end{bmatrix} \approx \left[\begin{array}{lll}0.924&0.383&0\\ -0.383&0.924&0\\ 0&0&1\end{array}\right]$$
This gives:
$$SMJ^{1/16}S^{-1} \approx \left[\begin{array}{lll}0.979&0.276&-0.165\\-0.165&0.979&0.276\\0.276&-0.165&0.979\end{array}\right] $$
Which we can verify is close to ${\bf A}^{16}$ if raised to 16.
Is my thinking sound?
Are there some obvious useful solutions which this approach will miss?
There can be infinitely many solutions. For example, $$ \pmatrix{s & t\cr (1-s^2)/t & - s}^2 = \pmatrix{1 & 0\cr 0 & 1\cr} $$ for all $s$, $t$ with $t \ne 0$.