Find $P$ such that $P^TAP = D$ where $D$ is a diagonal matrix. 
Let $$A = \left(\begin{array}{cc} 2&3 \\ 3&4  \end{array}\right) \in
M_n(\mathbb{C})$$
Find $P$ such that $P^TAP = D$ where $D$ is a diagonal matrix.

How can I find $P$? I am doing Gauss but it does not work?$$A = \left(\begin{array}{cc|cc} 2&3&1&0\\ 3&4&0&1    \end{array}\right) \sim \left(\begin{array}{cc|cc} 2&0&-8&6\\ 0&-1/2&-3/2&1    \end{array}\right)$$
What am I doing wrong? Steps would be much appreciated.

You need to perform simulatenous row and column operations on the left hand side while performing only column operations on the right hand side. Then, when the left side becomes diagonal, the right side will be your $P$. In your case,
$$ \left(\begin{array}{cc|cc} 2&3&1&0\\ 3&4&0&1    \end{array}\right) \xrightarrow[C_2 = C_2 - \frac{3}{2}C_1]{R_2 = R_2 - \frac{3}{2}R_1} 
\left(\begin{array}{cc|cc} 2&0&1& -\frac{3}{2}\\ 0& -\frac{1}{2}&0&1    \end{array}\right) $$
and indeed
$$ \begin{pmatrix} 1 & 0 \\ -\frac{3}{2} & 1 \end{pmatrix} \begin{pmatrix} 2 & 3 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 1 & -\frac{3}{2} \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & -\frac{1}{2} \end{pmatrix}. $$
