I’m trying to redefine the P matrix so that it’s made of eigenvectors with integer components (which will give the same diagonalizing matrix). But I’m stuck because I’m not sure how to redefine all the columns. I think the first column will be redefined as 1,-1,1,-1. But for the rest of the columns, I'm stuck. Please help me out, if possible. Thank you. (Original Matrix -> H)
$$A = \begin{pmatrix} \frac{1}{2} & \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\ \end{pmatrix}$$
$$A = P D P^{-1} =\begin{pmatrix} -1 & 1 & -1 & 1 \\ 1 & 0 & 0 & 1 \\ -1 & 0 & 1 & 0 \\ 1 & 1 & 0 & 0 \\ \end{pmatrix} \begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}\begin{pmatrix} -\frac{1}{4} & \frac{1}{4} & -\frac{1}{4} & \frac{1}{4} \\ \frac{1}{4} & -\frac{1}{4} & \frac{1}{4} & \frac{3}{4} \\ -\frac{1}{4} & \frac{1}{4} & \frac{3}{4} & \frac{1}{4} \\ \frac{1}{4} & \frac{3}{4} & \frac{1}{4} & -\frac{1}{4} \\ \end{pmatrix}$$
Note: You can also factor our $\frac{1}{2}$ from $A$. This will change $D$ to just be two times what is shown, but $P$ and $P^{-1}$ will be the same.