$$A= \begin{pmatrix} 1 & 0 & 0 & 0\\ 2 & 3 & 2 & 2\\ 2 & 2 & 3 & 2\\ 2 & 2 & 2 & 3\\ \end{pmatrix}$$
I know that the eigenvalues are 1 of geometric multiplicity = algebric multiplicity = 3, and 7 of geometric multiplicity = algebric multiplicity = 1.
The eigenvectors of 1 are $(1, 0, 0, -1),(1, 0, -1, 0),(1, -1, 0, 0)$ and of 7 is $(0, 1, 1, 1)$.
I know that $A$ is diagonalizable and is similar to, for example:
$$D= \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 7\\ \end{pmatrix}$$
but how can I show that $A$ is similar to
$$D_1= \begin{pmatrix} 1 & 1 & 0 & 0\\ 0 & 7 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ \end{pmatrix}$$
but not to:
$$D_2= \begin{pmatrix} 1 & 1 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 7\\ \end{pmatrix}$$
I see that the difference is in the order we choose the basis that the eigenvectores span, but how does it make difference? Suddenly $A$ is similar to non-diagonal matrix? so what the difference between $D_1$ and $D_2$?
