I have the following task:
Let $A = \left( \begin{array}{ccc} 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \end{array} \right)$. Find an invertible matrix $P$ s.t. $P'=PA$ is symmetric. Corresponding, find an orthornormal matrix $Q$ (whose columns form an orthonormal basis for $\mathbb{R}^4$ with respect to the standard inner product) and a diagonal matrix $D$ s.t. $P'=Q^TDQ$
Until now I know that:
If I choose $P:= A^T$ then $PA$ is symmetric. But I don't see how $PA$ could equal to $Q^TDQ$, when I choose $Q$ as the identity matrix. Because if I choose Q as the identity matrix then I would have to prove that $PA$ = D. But this is impossible, right? Do I have to change my Q?
It is a theorem that any symmetric matrix $S$ is diagonalizable, i.e. we can find an invertible matrix $Q$ and a diagonal matrix $D$ such that $S=Q^{-1}DQ$. Furthermore, when $S$ is symmetric, we can choose $Q$ to be orthogonal (same as orthonormal).
This certainly does not mean that $Q$ is the identity matrix. You should read up on how to diagonalise a matrix. There is a specific procedure involving eigenvectors. Take care to choose orthonormal eigenvectors in the process.
(For example, $\begin{pmatrix}0&-1\\1&0\\\end{pmatrix}$ is orthogonal but not the identity matrix.)