# 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 $D_2$?

• $$A \not \sim D_1$$ – Kaster Oct 7 '15 at 18:40
• @Kaster why ......... ? – Stabilo Oct 7 '15 at 18:50
• Can you provide $B$ that satisfies $A = BD_1 B^{-1}$? – Kaster Oct 7 '15 at 19:11
• @Kaster I explain in my answer that D1 is diagonalizable and it has the same eigenvalues like D, so they both similar to the same diagonal matrix. – Stabilo Oct 7 '15 at 19:17
• I'm sorry. You're right. I used wrong matrix in the decomposition, which was non-diagonalizable. – Kaster Oct 7 '15 at 19:30

## 1 Answer

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.