MHB Eigenvalues of similar matrices

AI Thread Summary
The discussion centers on the eigenvalues of a symmetric matrix A expressed as A = R * diag(a1, a2, a3) * R^T, where R is an orthogonal rotation matrix. It is established that since A and D = diag(a1, a2, a3) are similar matrices, they share the same eigenvalues, which are a1, a2, and a3. The main problem involves determining the characteristic polynomial of A, specifically through the determinant equation Det(R * diag(a1, a2, a3) * R^T - xI) = 0. The discussion also explores the relationship between the eigenvalues and a new parameter x*, leading to questions about the values of x in relation to diag(a1, a2, a3). Ultimately, the eigenvalues of A remain a1, a2, and a3, despite potential transformations in the characteristic polynomial.
Fernando Revilla
Gold Member
MHB
Messages
631
Reaction score
0
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.
 
Mathematics news on Phys.org
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 $a_3$.
 
Fernando Revilla said:
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)
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Back
Top