Diagonalizable matrix is similar to non-diagonal matrix??? $$A=
        \begin{pmatrix}
        1 & 0 & 0 & 0\\
        2 & 3 & 2 & 2\\
        2 & 2 & 3 & 2\\
        2 & 2 & 2 & 3\\
        \end{pmatrix}
$$
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 $D_2$?

Well, I figured it out...
$D_1$ and $D_2$ have the same eigenvalues as $A$ with the same algebric multiplicity but $D_2$ isn't diagonalizable because geometric multiplicity $\ne$ algebric multiplicity for every eigenvalue.
