This is a follow-up question on this one: Connection between eigenvalues and eigenvectors of a matrix in different bases
Assume I have matrix $ B=\left( \begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{array} \right). $
It's eigenvalues are $-1,-1,1,1$ with corresponding eigenvectors $(0,-1,0,1)^T,(-1,0,1,0)^T,(0,1,0,1)^T,(1,0,1,0)^T$.
Now if I chose a different basis, e.g. $d_1=(2,1,0,0)^T,d_2=(0,0,2,1)^T,d_3=(2,-1,0,0)^T,d_4=(0,0,2,-1)^T$ it becomes: $ [B]_{\mathcal{D}}=\left( \begin{array}{cccc} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array} \right). $
Now, the eigenvalues stay the same (as they should) but I have difficulties with the transformation of the eigenvectors. When I calculate them with $\mathcal{D}^{-1} v$ I e.g. get for the eigenvalue -1 $(-0.5,0.5,0.5,-0.5)^T$ and $(-0.25,0.25,-0.25,0.25)^T$ instead of the correct $(0,0,-1,1)^T$ and $(-1,1,0,0)^T$. Both are obviously not multiples of each other.
Can anybody help - what is wrong here (it must be some careless mistake, I guess)? Thank you!
EDIT
There was a nasty typo in the first matrix - the last $1$ should have been a $0$ - now everything seems correct. I am really sorry!