Eigenvalues of similar matrices

[Fernando Revilla]

I quote a question from Yahoo! Answers

If A=[R]*[diag(a1,a2,a3)]*[R]t can we conclude that a1,a2,a3 are the eigenvalues of A?(R is a rotation matrix)?
- We now that A is symmetric.
- R is a rotation matrix so it is orthogonal.
- [R]t is the transpose of R.

I have given a link to the topic there so the OP can see my response.

 

[Fernando Revilla]

In general, if $A,B\in \mathbb{F}^{n\times n}$ are similar matrices then, $A$ and $B$ have the same characteristic polynomial, as a consequence the same eigenvalues. In our case we have:
$$A=R\text{ diag }(a_1,a_2,a_3)\;R^T=R\text{ diag }(a_1,a_2,a_3)\;R^{-1}$$
so, $A$ and $D=\text{diag }(a_1,a_2,a_3)$ are similar matrices. But the eigenvalues of $D$ are $a_1$, $a_2$ and $a_3$, hence the eigenvalues of $A$ are also $a_1$, $a_2$ and $a_3$.

 

[Abbas]

In general, if $A,B\in \mathbb{F}^{n\times n}$ are similar matrices then, $A$ and $B$ have the same characteristic polynomial, as a consequence the same eigenvalues. In our case we have:
$$A=R\text{ diag }(a_1,a_2,a_3)\;R^T=R\text{ diag }(a_1,a_2,a_3)\;R^{-1}$$
so, $A$ and $D=\text{diag }(a_1,a_2,a_3)$ are similar matrices. But the eigenvalues of $D$ are $a_1$, $a_2$ and $a_3$, hence the eigenvalues of $A$ are also $a_1$, $a_2$ and $a_3$.

Thank You. So just the sequence of eigenvalues changes. The main Problem is:
Det( diag(a₁,a₂,a₃) + R diag(a₁,a₂,a₃)R^T - xI )=0 I rewrote it in this form:
Det( R diag (a₁,a₂,a₃)R^T - x^*I )=0
which means eigenvalues of R diag(a₁,a₂,a₃)R^T are the new parameter x*=x-diag(a₁,a₂,a₃)
can we say x-diag(a₁,a₂,a₃)= diag(a₁,a₂,a₃) or x=2 diag(a₁,a₂,a₃)