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
 A: 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.
A: A good first method of attack for finding matrices whose square is a given matrix, is to diagonalise if possible. The matrix which you have given, which we may call $P'$, is symmetric. Therefore, not only is it diagonalisable, it is orthogonally diagonalisable! I.e. there is an orthogonal matrix $S$ $(S^{-1} = S^T)$, such that $$P' = S D S^{-1} = S D S^T, $$ where $D$ is diagonal. If we find a matrix $B$ such that $B^2 = D$, we can generate a solution by setting $P = S B S^{-1}$, since \begin{align*}P^2 &= S B S^{-1} S B S^{-1} \\
&= S B^2 S^{-1} \\
& = S D S^{-1} \\
&= P'. \end{align*}
Square rooting a diagonal matrix is nice and easy, since $$\begin{pmatrix} \pm\sqrt{a} & 0 \\ 0 & \pm\sqrt{b} \end{pmatrix}^2 = \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix},$$ there are four solutions.
Using this method, the obtained solutions for $P$ will be symmetric matrices, since in each case, $P$ has been written such that it is orthogonally diagonalisable, and any orthogonally diagonalisable matrix is necessarily symmetric (why?).
An interesting question is whether or not there are more solutions. It seems that there are none, but I'm not sure how to prove it. Perhaps I will edit my answer when or if I find out. Does anyone have any ideas?
A: What you have here are split-complex numbers. These are just numbers of the form $\,p =  a+j\,b\,$ where $\,j^2=1.\,$ They are isomorphic to matrices of the form $\,P=(^{a\, b}_{b\,a}).\,$
They are also isomorphic to pairs or real numbers $\,(a+b,a-b)\,$ with pairwise addition and multiplication operations.
 Thus,
 $$ p^2 = \, (a+j\,b)^2 = \, (b+j\,a)^2 = \,(a^2+b^2)+j(2ab).$$ Notice that there are four square roots in general. They are $\,a+j\,b,\;$ $-a-j\,b,\;$ $b+j\,a,\;$ $-b-j\,a\,$
unlike the case of complex numbers with $\,i^2=-1\,$ where only roots are $\,a+i\,b,\;$ and $\,-a-i\,b.$
In your case, $\,c+j\,d := p^2=0.6+j\,0.4.\,$ To take the square root you just take the square root of both $\,c+d\,$ and $\,c-d.\,$ Therefore define 
$$ u\!:=\!a \!+\! b \!=\!\sqrt{0.6 \!+\! 0.4} \!=\! 1.0\;\;
\textrm{ and }\;\;
 v \!:=\! a \!-\! b \!=\! \sqrt{0.6 \!-\! 0.4} \!=\! \sqrt{0.2}$$
 so that 
$$ \sqrt{c + j\,d} = p\;\; \textrm{ and } \;\; (\sqrt{c+d}, \sqrt{c-d}) =  (u, v) =  (a+b,a-b).$$ When combined together they give you one of the solutions
 $$\,p=a+j\, b = (u+v)/2+j\,(u-v)/2$$
and you also have the other three given above.
P.S. You may be wondering why there are four square roots for split-complex numbers. The reason is contained in the very definition where $\,j^2=1.\,$ That is, in this number system there are four square roots of unity, namely $\,1,\,-1,\,j,\,-j,\,$ and hence also for any non-zero number. This is in contrast to complex numbers where $\,i^2=-1\,$ and the fourth roots of unity are $\,1,\,-1,\,i,\,-i.$
A: Let's say that your original matrix is called $A$. There are two obvious eigenvectors, $v_1 = (1, 1)$ with eigenvalue $\lambda_1 = 1$, and $v_2 = (1, -1)$ with eigenvalue $\lambda_2 = 0.2$.
Now, suppose that $P$ is an operator satisfying $P^2 = A$. Then in particular, $P$ and $A$ commute, since $P^3 = AP = PA$, and hence preserve each other's eigenspaces. Since the eigenspaces of $A$ are one-dimensional, this means that any solution $P$ will be of the form $Pv_1 = \mu_1 v_1$ and $Pv_2 = \mu_2 v_2$ for scalars $\mu_1, \mu_2$ satsifying $\mu_1^2 = \lambda_1$ and $\mu_2^2 = \lambda_2$. There are four obvious choices here, since $\mu_1 \in \{1, -1\}$ and $\mu_2 \in \{\sqrt{0.2}, -\sqrt{0.2}\}$.
So each solution may be found by picking one of the choices $(\mu_1, \mu_2)$ above, and writing down the corresponding matrix.
