<s>[INST] If $A$ and $B$ are positive-definite matrices with $A>B$, must $A^k > B^k$ for $k>0$? Assume $A$ and $B$ are positive-definite matrices. If $A>B$, can we conclude that $A^k > B^k$ for any positive scalar $k$?
Note that $A>B$ means $A-B$ is a positive-definite matrix, not an element-wise comparison. I've tried using Cayley-Hamilton theorem but could not get anywhere with it so far.
 [/INST] No. In $\mathbb{R}^{2 \times 2}$, take $B=I$ and
$$ A = \left( \begin{array}{cc} 2 & -2 \\ 2 & 2 \end{array} \right) $$
$A$ is positive-definite since
$$ (\begin{array}{cc} x & y \end{array})
 \left( \begin{array}{cc} 2 & -2 \\ 2 & 2 \end{array} \right)
\left( \begin{array}{c} x \\ y \end{array} \right) = 2 x^2 + 2 y^2 $$
$A-B$ is positive-definite since
$$ (\begin{array}{cc} x & y \end{array})
 \left( \begin{array}{cc} 1 & -2 \\ 2 & 1 \end{array} \right)
\left( \begin{array}{c} x \\ y \end{array} \right) = x^2 + y^2 $$
But
$$ A^2-B^2 = \left( \begin{array}{cc} 0 & -8 \\ 8 & 0 \end{array} \right) - I
= \left( \begin{array}{cc} -1 & -8 \\ 8 & -1 \end{array} \right) $$
and
$$ (\begin{array}{cc} 1 & 0 \end{array})
 \left( \begin{array}{cc} -1 & -8 \\ 8 & -1 \end{array} \right)
\left( \begin{array}{c} 1 \\ 0 \end{array} \right) = -1 < 0$$
</s>