Find a matrix $P$ such that $P^T H P$ is a diagonal matrix with non integer eigenvalues I have this problem in my textbook:
Find a matrix $P$ such that $P^T H P$ is a diagonal matrix, where $H :=  \begin{bmatrix}2 & 1 \\ 1 & 3\end{bmatrix}$
My attempt:
$$\det(A-\lambda I) = 6 - 5\lambda + \lambda ^2 -1$$
$$= 5 - 5 \lambda + \lambda ^2 $$
$$\therefore \lambda \in \left\{\frac{1}{2}(5-\sqrt{5}),\frac{1}{2}(5+\sqrt{5}) \right\}$$
Eigenvector for $\frac{1}{2}(5-\sqrt{5}):$
???
Eigenvector for $\frac{1}{2}(5+\sqrt{5}):$
???
$$\therefore \text{ let } P = \text{ ???} $$
Check: $P^tAP$
Something to show diagonal entries are the eigenvalues as expected.
I am having difficulty with the calculations with the non integer eigenvalues basically.
EDIT:
Thanks Thomas but what about finding $P$ I know there is a formula to find what goes in front of the matrix that comes from the eigenvectors (what do I put as $\frac{X}{Y}$ in $P=\frac{X}{Y}\begin{bmatrix}-1-\sqrt5 & 1+\sqrt5\\2 & 2 \end{bmatrix} $ for it to make sense

Just do the normal procedure of finding eigenvectors for the complex numbers.
$$\left[\begin{array}{cc|c}2-\frac{5-\sqrt5}{2}&1&0\\1&3-\frac{5-\sqrt5}{2}&0\end{array}\right]\Rightarrow\left[\begin{array}{cc|c}4-5+\sqrt5&2&0\\2&6-5+\sqrt5&0\end{array}\right]\Rightarrow\left[\begin{array}{cc|c}-1+\sqrt5&2&0\\2&1+\sqrt5&0\end{array}\right]$$Multiply row 1 by $-1-\sqrt5$,
$$\left[\begin{array}{cc|c}-4&-2-2\sqrt5&0\\2&1+\sqrt5&0\end{array}\right]\Rightarrow\left[\begin{array}{cc|c}-2&-1-\sqrt5&0\\2&1+\sqrt5&0\end{array}\right]\Rightarrow\left[\begin{array}{cc|c}2&1+\sqrt5&0\\0&0&0\end{array}\right]$$Thus $v_1=\begin{bmatrix}-1-\sqrt5\\2\end{bmatrix}$, correspondingly $v_2=\begin{bmatrix}-1+\sqrt5\\2\end{bmatrix}$
Now you can finish the rest of the problem.
