can this be de-coupled?

Question

irishboy123 · Answer

$ \mathbf{\dot y} = \left[ \begin{matrix} 4 & 2 \  2 & 1 \end{matrix} \right] \mathbf{y} $


one of the e-v's is zero, is that where the problem is??

ikram002p · Answer

it can be de-coupled as the eigenvalues are distinct reals, so the eigenvectors matrix  is diagonalizable  so we could find it's inverse which D.

det[A]=0 is not a problem. we only care for E matrices determinant to not being 0.

empty · Answer

@ikram002p is right, the fact that it has a zero eigenvalue isn't a problem, you can solve this just like any other. The only consequence of the eigenvalue being zero is that you will have a constant solution to your differential equation.

Solve for the entries of the eigenvector:
 $ \left[ \begin{matrix} 4 & 2 \  2 & 1 \end{matrix} \right]  \left[ \begin{matrix} a \  b\end{matrix} \right]   = 0 \left[ \begin{matrix} a \  b\end{matrix} \right]   $
Either equation ends up in the same equation, 
$$2a+b=0$$
So we can plug this into our eigenvector, and we can normalize it too in case it ends up being orthogonal to the other one (Which is always true for symmetric matrices with nonzero determinants, there's a cute simple proof of this fact if you don't know it I can show you!)

$\frac{1}{\sqrt{5}}\left[ \begin{matrix} 1 \  -2\end{matrix} \right]   $
It turns out our other eigenvector with eigenvalue 5 is orthogonal to it, 
$\frac{1}{\sqrt{5}}\left[ \begin{matrix} 2 \  1\end{matrix} \right]   $

That's great cause that means it's easy to invert the matrix of eigenvalues, it's just the transpose since every entry of a matrix multiplication comes from a dot product of a row with a column vector, since the eigenvectors that make it up are orthonormal only the diagonal exists and it's 1 instead of 5s cause we divided it out when we normalized them.

$AE=ED$ This matrix equation essentially encodes all the eigenvector equations into one, and it also is how we can decouple the equations.

$$\dot y = A y$$
$$\dot y = EDE^\top y$$
$$E^\top \dot y = DE^\top y$$
Now we can see that since the $E^\top$ matrix is constant that we can pull it inside the derivative and we now have successfully decoupled it:

$$\frac{d}{dt}\left( E^\top y \right) = D \left( E^\top y \right) $$
because we have the linear transformation:

$$u = E^\top y$$

which gives us:

$$\dot u = D u$$
Or if we look specifically at the end two equations:

$$\dot u_1 = 0$$$$\dot u_2 = 5 u_2$$

You can solve these and then transform it back, although I am not sure if this is exactly what you mean by decouple I don't solve these often. Oh and btw the E matrix I never gave it is literally the eigenvectors like I described:

$$E = \frac{1}{\sqrt{5}} \left[ \begin{matrix} 1 & 2 \ -2 & 1 \end{matrix} \right] $$

irishboy123 · Answer

Thanks mucho @Empty, nail hit on head.

I made those numbers up a bit, but the real working example I have to play with asks me to de-couple 
$S[y_1, y_2] = \int  \; y_1'^2 + y_2'^2 - (y_1 + 3 y_2)^2 \; dx$ 
using linear transforms 
$z_1 = y_1 + 3 y_2, \; z_2 = a y_1 + y_2$.

I can't face the algebra, that's how you are supposed to do it, so I already guessed that the "neutral" lines for that system lie along the e-vectors.

The E-L equation for the system gives out 
$\mathbf {y''} = \left( \begin{matrix}
  -1 & -3 \
  -3 & -9
\end{matrix}
 \right) \mathbf {y}$

The e-values are 0 and -10 and your E matrix is 
$\mathbf {E} = \left( \begin{matrix}
  1 & -3 \
  3 & 1
\end{matrix}
 \right) $
So
$\mathbf {E^{-1}} = {1 \over 10} \left( \begin{matrix}
  1 & 3 \
  -3 & 1
\end{matrix}
 \right) $

Which magically nails this in a few lines.  Awepik!

a = -3......which i know to be correct from seeing it done the long way.


That gives me $z_1'' = - 10 z_1, z_2 '' = 0$ so that proves the de-coupling without my having to consider the matrix algebra/proof stuff too, i think.

I thank you.

irishboy123 · Answer

and thank you @ikram002p !  det(E) = 10.

ikram002p · Answer

=D