<s>[INST] Developing a general expression for powers of matrices Is there a relationship for matrices that are squared? 
I am trying to determine two possible matrices $P$ where
$$P^2=\begin{bmatrix}0.6&0.4\\0.4&0.6\end{bmatrix}$$
I know that $P$ has to be a $2×2$ matrix and that $P$ will also be symmetric.
Thanks
 [/INST] If you assume that $$P=\begin{pmatrix}a&b\\b&c\end{pmatrix}$$
then you compute $$P^2=\begin{pmatrix}a^2+b^2&ab+bc\\ab+bc&b^2+c^2\end{pmatrix}=\begin{pmatrix}0.6&0.4\\0.4&0.6\end{pmatrix}$$
From this, you get the system of equations $$a^2+b^2=0.6,~ b(a+c)=0.4,~ b^2+c^2=0.6$$
Subtracting the third from the first, we get $a^2-c^2=0$, or $a=\pm c$.  $a=-c$ is inconsistent with the second equation, so we must have $c=a$, and now two equations: $$a^2+b^2=0.6, 2ab=0.4$$
Adding, we get $a^2+2ab+b^2=(a+b)^2=1$, so $a+b=\pm 1$.  Also, subtracting we get $a^2-2ab+b^2=(a-b)^2=0.2$, so $a-b=\pm \sqrt{0.2}$.  Hence, there are four cases, corresponding to $$a+b=1, a-b=\sqrt{0.2}$$ $$a+b=-1, a-b=\sqrt{0.2}$$ $$a+b=1, a-b=-\sqrt{0.2}$$ $$a+b=-1, a-b=-\sqrt{0.2}$$
The first gives $a\approx 0.72, b\approx 0.28$.  The second gives $a\approx -0.28, b\approx -0.72$.  The third gives $a\approx 0.28, b\approx 0.72$.  The fourth gives $a\approx -0.72, b\approx -0.28$.  Hence there are just four answers, two of which are the negatives of the other two.
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