$A^{2}-2A=\begin{bmatrix} 5 & -6 \\ -4 & 2 \end{bmatrix}$
Can someone help me solve this? I've been trying to solve it for a while, but no matter what I try, the only information that I manage to get about A is that if $A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ then $c=\frac{2b}{3}$. Any help would be appreciated, thanks!
On
$$A^{2}-2A+I=\begin{bmatrix} 5 & -6 \\ -4 & 2 \end{bmatrix}+I \\ (A-I)^2=\begin{bmatrix} 6 & -6 \\ -4 & 3 \end{bmatrix} $$
We know that, if$$\begin{align}M&=PDP^{-1} \\ M^n&=PD^nP^{-1}\end{align}$$
Let $B=A-I$ then, $$B=\sqrt{\begin{bmatrix} 6 & -6 \\ -4 & 3 \end{bmatrix}}$$
Diagonalise $B^2$ as, $$\left(\begin{matrix} \frac{\sqrt{105}-3}{8} & \frac{-\sqrt{105}-3}{8} \\ 1 & 1 \end{matrix}\right).\left(\begin{matrix} \frac{-\sqrt{105}+9}{2} & 0 \\ 0 & \frac{\sqrt{105}+9}{2} \end{matrix}\right).\left(\begin{matrix} \frac{4*\sqrt{105}}{105} & \frac{\sqrt{105}+35}{70} \\ \frac{-4*\sqrt{105}}{105} & \frac{-\sqrt{105}+35}{70} \end{matrix}\right)$$
From this, $$B=\left(\begin{matrix} \frac{\sqrt{105}-3}{8} & \frac{-\sqrt{105}-3}{8} \\ 1 & 1 \end{matrix}\right).\left(\begin{matrix} \frac{-\sqrt{105}+9}{2} & 0 \\ 0 & \frac{\sqrt{105}+9}{2} \end{matrix}\right)^{\frac{1}{2}}.\left(\begin{matrix} \frac{4*\sqrt{105}}{105} & \frac{\sqrt{105}+35}{70} \\ \frac{-4*\sqrt{105}}{105} & \frac{-\sqrt{105}+35}{70} \end{matrix}\right)$$
From this $$A=\left(\begin{matrix} \frac{\sqrt{105}-3}{8} & \frac{-\sqrt{105}-3}{8} \\ 1 & 1 \end{matrix}\right).\left(\begin{matrix} \frac{-\sqrt{105}+9}{2} & 0 \\ 0 & \frac{\sqrt{105}+9}{2} \end{matrix}\right)^{\frac{1}{2}}.\left(\begin{matrix} \frac{4*\sqrt{105}}{105} & \frac{\sqrt{105}+35}{70} \\ \frac{-4*\sqrt{105}}{105} & \frac{-\sqrt{105}+35}{70} \end{matrix}\right)+\left(\begin{matrix} 1&0\\0&1\end{matrix}\right)$$
The required matrix is approximately $$\tiny{\left(\begin{matrix} \frac{\left(481173769149-31173769149*i\right)*\sqrt{105}+\left(20341081920215+1091081920215*i\right)}{3500000000000} & \frac{\left(-481173769149+31173769149*i\right)*\sqrt{105}+875000000000}{875000000000} \\ \frac{\left(-160391256383+10391256383*i\right)*\sqrt{105}}{437500000000} & \frac{\left(-481173769149+31173769149*i\right)*\sqrt{105}+\left(20341081920215+1091081920215*i\right)}{3500000000000} \end{matrix}\right)}$$
On
$\newcommand{\Tr}{\mathrm{Tr}\,}$ Let $Y=A-I$, $X=B+I$, $B$ for the rhs matrix. Then $\Tr X = 9$, $\det X=-6$, and we need to solve $$Y^2=X$$ for $Y$. Write $\alpha=\Tr Y$, $\beta = \det Y$. Then $$Y^2-\alpha Y + \beta I=0$$ or $$\alpha Y = \beta I + X$$ so that finding allowed values of $\alpha$, $\beta$ solves the problem.
Take the trace and determinant of both sides of $Y^2=X$. Then $$\begin{align} \alpha^2 - 2\beta &= \Tr X = 9 \\ \beta^2 &= \det X = -6 \end{align}$$ which means that $$\begin{align} \alpha A &= (\alpha+\beta +1)I + B \\ \alpha^2 & = 9 +2\beta \\ \beta^2 &= -6\text{.} \end{align}$$ Note that there are four solutions.
Denote by $I$ the identity matrix. Then, completing squares you can write $$A^2 - 2A = A^2 -2IA + I^2 -I^2 = (A-I)^2 -I^2.$$ Hence, your equation is equivalent to $$(A-I)^2 = X + I$$ since $I^2 = I$. Denote by $Y=X+I$ the new matrix (which is known). You want to find $B$ such that $B^2=Y.$ Here, I recommend to diagonalize $Y$, i.e. find $U$ and diagonal $D$ such that $$Y=UDU^{-1}.$$ Thus, $$B= Y^{1/2} = UD^{1/2}U^{-1}.$$ See this link for more information.
Once you have found $B$, $A$ is given by $$A=B+I.$$ Remember that you may have more than one square root of the matrix $Y$.