I was trying to diagonalize a not really nice matrix doing first a change of basis but I noticed that the two characteristic polynomials I get are different.
Original matrix and its characteristic polynomial is:
$ M_{\alpha} =\begin{bmatrix} 1 & -1 & 0 & -1 \\ \alpha & -\alpha+1 & 1 & -\alpha \\ \alpha & \alpha+1 & 2\alpha+1 & -\alpha \\ 1 & -1 & 0 & -1 \end{bmatrix} $
$p_a(\lambda)=\lambda^4-(2+a) \lambda^3 -2 a^2 \lambda^2$
My new basis:
$\mathcal{B'}=\{(1,\alpha, 0, 0), (0,1,1,0), (0,0,\alpha,0), (0,0,0,1) \}$
According to this basis $M_\alpha$ should be similar to:
$ M'_{\alpha} =\begin{bmatrix} 1 & -1 & 0 & -1 \\ 0 & 1 & 1 & 0 \\ 1 & 1 & 2 & -1 \\ 1 & -1 & 0 & -1 \end{bmatrix} $
and its characteristic polynomial should be: $p'(\lambda) = \lambda^4-3\lambda^2+\lambda^2 $
What am I doing wrong? (In any case I don't think my change of basis is the smartest one… something else?)
A matrix represents a map from one space to another (in this case $\mathbb{R}^4$ to $\mathbb{R}^4$). It is written with respect to two bases, one for each space. In this case, you want to use the same basis for both spaces.
However, to get from your first matrix to the second, you have only changed one of the two bases. To get the correct new matrix, you should multiply the original matrix by the change of basis matrix on one side, and by its inverse on the other side.