1
$\begingroup$

$$A= \begin{pmatrix} 1 & 0 & 0 & 0\\ 2 & 3 & 2 & 2\\ 2 & 2 & 3 & 2\\ 2 & 2 & 2 & 3\\ \end{pmatrix} $$

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 $D_2$?

$\endgroup$
  • $\begingroup$ $$ A \not \sim D_1 $$ $\endgroup$ – Kaster Oct 7 '15 at 18:40
  • $\begingroup$ @Kaster why ......... ? $\endgroup$ – Stabilo Oct 7 '15 at 18:50
  • $\begingroup$ Can you provide $B$ that satisfies $A = BD_1 B^{-1}$? $\endgroup$ – Kaster Oct 7 '15 at 19:11
  • $\begingroup$ @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. $\endgroup$ – Stabilo Oct 7 '15 at 19:17
  • $\begingroup$ I'm sorry. You're right. I used wrong matrix in the decomposition, which was non-diagonalizable. $\endgroup$ – Kaster Oct 7 '15 at 19:30
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.