# Similar to an eigenvalue problem . . . help!

1. Nov 28, 2009

### BobbyBear

Consider the following linear homogeneous ordinary differential equation system:
(NB this system describes the movement of the natural response of a two degree of freedom structural system made up of two lumped masses connected by elastic rigidities) :

$$\left( \begin{array}{cc} m_1 & 0 \\ 0 & m_2 \\ \end{array} \right) \left( \begin{array}{cc} \ddot{u}_1 \\ \ddot{u}_2 \\ \end{array} \right) + \left( \begin{array}{cc} (k_1 + k_2) & -k_2 \\ -k_2 & k_2 \\ \end{array} \right) \left( \begin{array}{cc} u_1 \\ u_2 \\ \end{array} \right) = \left( \begin{array}{cc} 0 \\ 0 \\ \end{array} \right)$$

which I shall compactly write as:

$$[m] \vec{\ddot{u}} + [k] \vec{u}} = \vec{0}$$

Now, to solve, we assume a solution of the form:

$$\vec{u}(t)=q_n(t) \vec{\phi _n}$$

where

$$q_n(t) = A_n cos (\omega _n t) + B_n sin (\omega _n t)$$

and

$$\vec{\phi _n}$$

is a constant vector.

Then
$$\vec{\ddot{u}}(t)=-\omega _n^2 q_n(t) \vec{\phi _n}$$

Substituting into the differential system,

$$\left[-\omega _n^2 [m] \vec{\phi _n} + [k] \vec{\phi _n} \right] q_n(t) = \vec{0}$$

from which

$$-\omega _n^2 [m] \vec{\phi _n} + [k] \vec{\phi _n} = \vec{0}$$

$$(-\omega _n^2 [m] + [k]) \vec{\phi _n} = \vec{0}$$

and for there to be a non trivial solution, we need:

$$det(-\omega _n^2 [m] + [k]) = 0$$

from which we get two values of

$$\omega _n^2$$

Now, my book (Dynamics of Structures by Chopra) says that the $$\omega _n^2$$ are real and positive because [k] and [m] are real symmetric and positive definite.
I don't see how this deduction is made!! I mean, I know that if a matrix [A] is a real symmetric matrix that is positive definite, then all its eigenvalues are real and positive (the proof is available in any standard text of linear algebra).
But I just don't see how to prove the other statement! the $$\omega _n^2$$ are not the eigenvalues of any matrix, are they? (even though it's a similar problem to an eigenvalue problem). Can someone help me see how that deduction is made?

2. Nov 29, 2009

### Mute

The matrix $[m]$ is invertible. Factor it out of

$$-\omega _n^2 [m] \vec{\phi _n} + [k] \vec{\phi _n} = \vec{0}$$

to get

$$-\omega _n^2 \vec{\phi _n} + [m]^{-1}[k] \vec{\phi _n} = \vec{0}$$

or

$$[m]^{-1}[k] \vec{\phi _n} = \omega _n^2 \vec{\phi _n},$$

which means the $\omega_n^2$ are eigenvalues of the matrix [m]^{-1}[k].

3. Nov 30, 2009

### BobbyBear

Thank you Mute, I never thought of doing that!

But I'm still not quite able to reach the desired conclusion...

Okay so the $\omega_n^2$ are eigenvalues of the matrix $[m]^{-1}[k]$

And I've read that every positive definite matrix is invertible, and its inverse is also positive definite, so that means that if [m] is positive definite, then so is $[m]^{-1}$

So we have that both $[m]^{-1}$ and $[k]$ are positive definite, but in general that does not mean that $[m]^{-1} [k]$ is positive definite, does it?
I've read that if two matrices [M] and [N] are positive definite, then their product is positive definite if $[M] [N] = [N] [M]$, but this is not the case with $[m]^{-1}$ and [k], their product is not commutative in general. So how can we see that $[m]^{-1} [k]$ is positive definite?

Oh but wait! I just realised that even though [m] and [k] are real symmetric, $[m]^{-1}[k]$ is not even symmetric, so it wouldn't be of any use to prove that $[m]^{-1}[k]$ is positive definite, would it, because we don't know that the eigenvalues are real...
$$[m]^{-1}[k] = \left( \begin{array}{cc} 1/m_1 & 0 \\ 0 & 1/m_2 \\ \end{array} \right) \left( \begin{array}{cc} (k_1 + k_2) & -k_2 \\ -k_2 & k_2 \\ \end{array} \right) = \left( \begin{array}{cc} (k_1 + k_2)/m_1 & -k_2/m_1 \\ -k_2/m_2 & k_2/m_2 \\ \end{array} \right)$$
So how do we see that the eigenvalues of $[m]^{-1}[k]$ are real and positive?