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Eigenvalues of similar matrices

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
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

Well-known member
MHB Math Helper
Jan 29, 2012
661
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

New member
Aug 9, 2013
4
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)RT - xI )=0 I rewrote it in this form:
Det( R diag (a1,a2,a3)RT - x*I )=0
which means eigenvalues of R diag(a1,a2,a3)RT 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)