I want to diagonalize the following matrix: $$A=\begin{bmatrix}1 & -4 \\ -4 & -5\end{bmatrix}$$
I computed its eigenvalues, $\lambda_1=3$ and $\lambda_2=-7$, and got that the eigenspace corresponding to $\lambda_1$ is $Span \left\{\begin{bmatrix}-2\\1\end{bmatrix}\right\}$, while that for $\lambda_2$ is $Span\left\{\begin{bmatrix}1/2 \\ 1\end{bmatrix}\right\}$.
My understanding is that the columns of $P$ in $A=PDP^T$ need only be orthogonal, so I can take any two eigenvectors from the two eigenspaces to form it, for example: $\mathbf{v}_1=\begin{bmatrix}-2 \\ 1\end{bmatrix}$ and $\mathbf{v}_2=\begin{bmatrix}1 \\ 2\end{bmatrix}$. So:
$$P=\begin{bmatrix}\mathbf{v}_1 & \mathbf{v}_2\end{bmatrix}=\begin{bmatrix}-2 & 1 \\ 1 & 2\end{bmatrix}\quad D = \begin{bmatrix}3 & 0 \\ 0 & -7\end{bmatrix}$$
However, when I checked if $A=PDP^T$, I got that:
$$PD=\begin{bmatrix}-2 & 1 \\ 1 & 2\end{bmatrix}\times\begin{bmatrix}3 & 0 \\ 0 & -7\end{bmatrix}=\begin{bmatrix}-6 &-7\\ 3 & -14\end{bmatrix}$$ and $$(PD)P^T=\begin{bmatrix}-6 &-7\\ 3 & -14\end{bmatrix}\times\begin{bmatrix}-2 & 1 \\ 1 & 2\end{bmatrix}=\begin{bmatrix}5 & -20 \\ -20 & -25\end{bmatrix}=5\begin{bmatrix}1 & -4 \\ -4 & -5\end{bmatrix}=5A$$
I guess that I am doing something wrong because the definition in my book (Linear Algebra and its Applications) mentions only orthogonality of $P$'s columns. However, if I normalize $\mathbf{v}_1$ and $\mathbf{v}_2$, i.e. construct $P$ as follows:
$$P=\begin{bmatrix}\frac{1}{\lVert \mathbf{v}_1 \rVert}\mathbf{v}_1 & \frac{1}{\lVert \mathbf{v}_2 \rVert}\mathbf{v}_2\end{bmatrix}$$
then it really holds that $PDP^T=A$.
So, do columns of $P$ in orthogonal decomposition have to be unit vectors? If not, what am I doing wrong?
The problem lies in the fect that you should be dealing with $PDP^{-1}$, which is not the same thing as $PDP^T$, although they are equal if $P^{-1}=P^T$, which happens if and only if $P$ is an orthogonal matrix.