Find the eigenvalues of a 3x3 matrix

In summary: I think I misunderstood.In summary, the person is asking for help finding the eigenvalues of a 3x3 matrix. They find the eigenvalues using quadratic formula and then solve for x.
  • #1
happyparticle
400
20
Homework Statement
Find the eigenvalues of a 3x3 matrix
Relevant Equations
##\begin{pmatrix} - \omega^2 +\frac{k}{m} &-\frac{k}{m} &0\\ -\frac{k}{M}& - \omega^2 + 2 \frac{k}{M} &-\frac{k}{M} \\
0&- \frac{k}{m} & -\omega^2 + \frac{k}{m}
\end{pmatrix} \begin{pmatrix}A_1\\A_2\\A_3 \end{pmatrix} =0##
Hi,
I have a 3 mass system. ##M \neq m##

I found the forces and I get the following matrix.

I have to find ##\omega_1 , \omega_2, \omega_3## I know I have to find the values of ##\omega## where det(A) = 0, but with a 3x3 matrix it is a nightmare. I can't find the values.

I'm wondering if there's another way to have the eigenvalues.
WxfG6.png

##\begin{pmatrix} - \omega^2 +\frac{k}{m} &-\frac{k}{m} &0\\ -\frac{k}{M}& - \omega^2 + 2 \frac{k}{M} &-\frac{k}{M} \\
0&- \frac{k}{m} & -\omega^2 + \frac{k}{m}
\end{pmatrix} \begin{pmatrix}A_1\\A_2\\A_3 \end{pmatrix} =0
##
 
Physics news on Phys.org
  • #2
It is not a nightmare. Set ##W=-\omega^2-q-x\, , \,V=W-q\, , \,q=-k/M## and develop the determinant along the first row. It took me eight lines to find the eigenvalues.
 
  • #3
fresh_42 said:
It is not a nightmare.
It's perhaps the sort of thing some people have nightmares about!
 
  • #4
fresh_42 said:
It is not a nightmare. Set ##W=-\omega^2-q-x\, , \,V=W-q\, , \,q=-k/M## and develop the determinant along the first row. It took me eight lines to find the eigenvalues.

I'm not sure to understand the way you set the variables. Why you have ##x## and why not ##W = \omega^2 + \frac{k}{m}##
 
  • #5
The eigenvalues are the roots of the polynomial ##\det (A-x\cdot I).## Hence we have ##a_{ii}-x## on the diagonal. You can also work with ##W-x## instead. I simply found it more convenient to calculate ##\det\left(\begin{bmatrix}
W&q&0\\q&V&q\\0&q&W
\end{bmatrix}\right) ## and replace the actual values afterwards rather than
##\det\left(\begin{bmatrix}
W-x&q&0\\q&W-q-x&q\\0&q&W-x
\end{bmatrix}\right) ##. It's a matter of taste.
 
  • #6
You typed ##q = \frac{k}{M}##, but you also replaced ##\frac{k}{m}## by q even if ##M \neq m##. I don't know if it matters.

At the end I have ##VW^2 - \frac{2k^2W}{mM}## = 0. I didn't replace ##\frac{k}{m}##

Using quadratic formula I have ##W = 0 ## and ##W = -4k^2/mM##

then
##-\omega^2 -\frac{k}{m} - x = \frac{-4k^2}{mM}##

It Doesn't give me the right eigenvalues which are, ##\omega^2= 0, k/m , k/m + 2k/M##

As usual I probably just don't understood correctly the explanation.
 
  • #7
You have to solve for the determinant ##\det(A-xI)## not just ##\det A##.
EpselonZero said:
You typed ##q = \frac{k}{M}##, but you also replaced ##\frac{k}{m}## by q even if ##M \neq m##. I don't know if it matters.

At the end I have ##VW^2 - \frac{2k^2W}{mM}## = 0. I didn't replace ##\frac{k}{m}##

Using quadratic formula I have ##W = 0 ## and ##W = -4k^2/mM##

then
##-\omega^2 -\frac{k}{m} - x = \frac{-4k^2}{mM}##

It Doesn't give me the right eigenvalues which are, ##\omega^2= 0, k/m , k/m + 2k/M##

As usual I probably just don't understood correctly the explanation.
You are right. I made a mistake as I used only one mass ##M=m##. Sorry.

My corrected result is ##\det (A-xI)= \ldots##
$$
0=W(VW-2qQ)=\left(-\omega^2+\dfrac{k}{m}-x\right)\cdot\left(\left(-\omega^2+\dfrac{k}{m}-x\right)\cdot \left(-\omega^2+2\dfrac{k}{M}-x\right)-2\dfrac{k^2}{mM}\right)
$$
which results in ##x=-\omega^2+\dfrac{k}{m}## and the quadratic equation
$$
0=x^2+2x\left(\omega^2-\dfrac{1}{2}\cdot\dfrac{k}{m}-\dfrac{k}{M}\right)+\left(\omega^4-\omega^2\dfrac{k}{m}-2\omega^2\dfrac{k}{M}\right)
$$
which is easy to solve (if I haven't made a mistake again).

In case ##\omega ## was your variable, then I do not understand where the square comes from. But then your calculation was correct. I also get ##0\, , \,\dfrac{k}{m}\, , \,\dfrac{k}{m}+2\dfrac{k}{M}.##
 
Last edited:
  • Like
Likes happyparticle
  • #8
I may have misunderstood the problem. You said "I need to find the eigenvalues of a ##3\times 3## matrix" and then came the matrix ... or did you mean I need to solve ##\det (A-\omega I)=0## with eigenvalues ##\omega ##? I thought it was the first question and then we need a variable, say ##x## for the polynomial. The second possibility has ##\omega ## as variable, or ##\omega^2## in your case. But why ##\omega^2?##

So what is the matrix you want to find the eigenvalues to?
Or, what is the cubic polynomial you want to find the roots to?
 
  • #9
fresh_42 said:
I may have misunderstood the problem. You said "I need to find the eigenvalues of a ##3\times 3## matrix" and then came the matrix ... or did you mean I need to solve ##\det (A-\omega I)=0## with eigenvalues ##\omega ##? I thought it was the first question and then we need a variable, say ##x## for the polynomial. The second possibility has ##\omega ## as variable, or ##\omega^2## in your case. But why ##\omega^2?##

So what is the matrix you want to find the eigenvalues to?
Or, what is the cubic polynomial you want to find the roots to?
Sorry I found my errors. You explanation was perfect. I just didn't realize I don't need to go that far to find the roots for ##\omega##

Thanks!
 

1. What is an eigenvalue?

An eigenvalue is a scalar value that represents the amount by which a matrix stretches or compresses a vector when multiplied by it. It is an important concept in linear algebra and is used to solve systems of equations and analyze the behavior of dynamic systems.

2. How do you find the eigenvalues of a 3x3 matrix?

To find the eigenvalues of a 3x3 matrix, you first need to calculate the determinant of the matrix. Then, you need to solve the characteristic equation, which is obtained by setting the determinant equal to 0. The solutions to the characteristic equation are the eigenvalues of the matrix.

3. Can a 3x3 matrix have complex eigenvalues?

Yes, a 3x3 matrix can have complex eigenvalues. This happens when the determinant of the matrix is negative and the solutions to the characteristic equation are complex numbers.

4. What is the significance of eigenvalues in matrix calculations?

Eigenvalues are important in matrix calculations because they provide information about the behavior of the matrix when it is multiplied by a vector. They also help in solving systems of linear equations and analyzing the stability of dynamic systems.

5. Is there a shortcut to finding the eigenvalues of a 3x3 matrix?

Yes, there is a shortcut called the trace-determinant method. It involves calculating the sum of the diagonal elements (trace) and the determinant of the matrix. This method can only be used for 3x3 matrices and provides a quicker way to find the eigenvalues.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
995
  • Introductory Physics Homework Help
Replies
3
Views
950
  • Introductory Physics Homework Help
Replies
1
Views
691
  • Introductory Physics Homework Help
Replies
6
Views
227
  • Advanced Physics Homework Help
Replies
13
Views
1K
  • Introductory Physics Homework Help
Replies
17
Views
373
  • Introductory Physics Homework Help
Replies
8
Views
334
  • Introductory Physics Homework Help
Replies
1
Views
804
  • Introductory Physics Homework Help
Replies
3
Views
215
  • Introductory Physics Homework Help
Replies
4
Views
737
Back
Top