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

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 : colorado.edu/physics/phys3210/phys3210_sp14/…
– user163104
Jul 23 '17 at 3:08
• 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. Jul 23 '17 at 3:18
• @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?". Jul 23 '17 at 6:47
• @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
– user163104
Jul 23 '17 at 7:10