Similar matrix for two projections Let there be two matrices $A = \frac{1}{2} \begin{pmatrix}
        1 & 1\\
        1 & 1\\
        \end{pmatrix}$ and $B = \frac{1}{4} \begin{pmatrix}
        2 & 4\\
        1 & 2\\ \end{pmatrix}$ over the field $\mathbb{F}=\mathbb{Q}$. Find a invertible matrix $S$, s.t. $S^{-1} A S = B$.
I tried to solve this via multiplication with simple matrices but I can't seem to find such $S$. 
These two matrices have the property that $A^2 = A$ and $B^2 = B$. Is there any way to use this property to find the matrix $S$ (over image/kernel properties) ?
 A: An example is $S= \left( \begin{array}{cc} 1 & 0\\ 0&2 \end{array} \right)$.
EDIT: A way to find this matrix is: Take $S= \left( \begin{array}{cc} a & 0\\ b&c \end{array} \right)$. It's very important to take that 0 for simplify your work. (Generally, when you want to find a matrix with some properties, try to take as many zeros as you can, and if taking, let's say $n$ zeros, you find nothing, try to take $n-1$ zeros, and so on.) Then $S^{-1}= \frac{1}{ac} \left( \begin{array}{cc} c & 0\\ -b&a \end{array} \right)$. The ginven relation becomes:
$ \frac{1}{ac} \left( \begin{array}{cc} c & 0\\ -b&a \end{array} \right) \cdot  \left( \begin{array}{cc} 1 & 1\\ 1&1 \end{array} \right) \cdot \left( \begin{array}{cc} a & 0\\b&c \end{array} \right)= \left( \begin{array}{cc} 1 & 2\\ \frac{1}{2}&1 \end{array} \right)$, or $ \left( \begin{array}{cc} ac+bc & c^2\\ (a-b)(a+b)&(a-b)c \end{array} \right)=\left( \begin{array}{cc} ac & 2ac\\ \frac{ac}{2}&ac \end{array} \right)$.
From $ac+bc=ac$ we get that $bc=0$, so $b=0$ or $c=0$. Clearly, $c \neq 0$ (because $S$ is invertible), so $b=0$. From $c^2=2ac$ we get that $c=2a$. In this case, we have $(a-b)(a+b)=\frac{ac}{2}$ and $(a-b)c=ac$. So $S=\left( \begin{array}{cc} a & 0\\ 0&2a \end{array} \right)$.
