Why is it that a coupled mass-spring system will always produce a diagonalizable matrix?

Question

If you take a system like the one in the image, and you do the $y=x'$ trick to turn it into a first-order system of equations ($x_{1}$ or $x_{2}$ being the displacement of the mass $m_{1}$ or $m_{2}$ respectively from equilibrium), you'll get a $4 \times 4$ matrix that could have repeated eigenvalues but will always have 4 eigenvectors according to a professor of mine. However, I would like to see a proof.

A proof, in physics?? The text of your question is leading towards MathSE , based on the formal treatment and properties of matrices, as referred to here : http://www.colorado.edu/physics/phys3210/phys3210_sp14/lecnotes.2014-03-31.Intro_to_Coupled_Oscillators.html — , Jul 23 '17 at 03:08
This system has two degrees of freedom hence two normal modes. A standard procedure will lead to second order equations of motions in terms two by two matrices. These matrices are real and symmetric hence they can be orthogonaly diagonalized. I suppose that when transforming it to a system of first order equations the four by four matrices are still symmetric. — Diracology, Jul 23 '17 at 03:18
May be helpful my answers herein (1) : Eigenvalue equation for kinetic and potential energy and therein (2) : Can we let the lowest of n by equal (lenght and k) springs connected masses in equilibrium move in a siusoid way after giving the lowest a pull?. — Frobenius, Jul 23 '17 at 05:54
@Countto10 : It would be reasonable imo to ask though "what is the physical intuition/reasoning behind the eigenvectors and why then physically there should be 4?". — The_Sympathizer, Jul 23 '17 at 06:47
@mike4ty4 You are correct, and I was too flippant. I have upvoted the question and I take your point to emphasise that a physics viewpoint is often (always?) available and preferable. Thanks — , Jul 23 '17 at 07:10

Why is it that a coupled mass-spring system will always produce a diagonalizable matrix?

0 Answers0