Similar to an eigenvalue problem.help

[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) :

<br /> \left( \begin{array}{cc}<br /> m_1 &amp; 0 \\<br /> 0 &amp; m_2 \\<br /> \end{array} \right) <br /> <br /> \left( \begin{array}{cc}<br /> \ddot{u}_1 \\<br /> \ddot{u}_2 \\<br /> \end{array} \right) <br /> +<br /> \left( \begin{array}{cc}<br /> (k_1 + k_2) &amp; -k_2 \\<br /> -k_2 &amp; k_2 \\<br /> \end{array} \right) <br /> <br /> \left( \begin{array}{cc}<br /> u_1 \\<br /> u_2 \\<br /> \end{array} \right) <br /> <br /> =<br /> \left( \begin{array}{cc}<br /> 0 \\<br /> 0 \\<br /> \end{array} \right) <br /> <br /> <br />

which I shall compactly write as:

<br /> [m] \vec{\ddot{u}} + [k] \vec{u}} = \vec{0}<br />

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

<br /> \vec{u}(t)=q_n(t) \vec{\phi _n} <br />

where

<br /> q_n(t) = A_n cos (\omega _n t) + B_n sin (\omega _n t)<br />

and

<br /> \vec{\phi _n} <br />

is a constant vector.

Then
<br /> \vec{\ddot{u}}(t)=-\omega _n^2 q_n(t) \vec{\phi _n}<br />

Substituting into the differential system,

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

from which

<br /> -\omega _n^2 [m] \vec{\phi _n} + [k] \vec{\phi _n} = \vec{0}<br />

<br /> (-\omega _n^2 [m] + [k]) \vec{\phi _n} = \vec{0}<br />

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

<br /> det(-\omega _n^2 [m] + [k]) = 0<br />

from which we get two values of

<br /> \omega _n^2<br />

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?

 

[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].

 

[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 <br /> \omega_n^2<br /> are eigenvalues of the matrix <br /> [m]^{-1}[k]<br />

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 <br /> [m]^{-1}<br />

So we have that both <br /> [m]^{-1}<br /> and <br /> [k]<br /> are positive definite, but in general that does not mean that <br /> [m]^{-1} [k]<br /> 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 <br /> [M] [N] = [N] [M]<br />, but this is not the case with <br /> [m]^{-1}<br /> and [k], their product is not commutative in general. So how can we see that <br /> [m]^{-1} [k]<br /> is positive definite?

Thanks for your help!

 

[BobbyBear]

Oh but wait! I just realized 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] =<br /> <br /> \left( \begin{array}{cc}<br /> 1/m_1 &amp; 0 \\<br /> 0 &amp; 1/m_2 \\<br /> \end{array} \right) <br /> <br /> \left( \begin{array}{cc}<br /> (k_1 + k_2) &amp; -k_2 \\<br /> -k_2 &amp; k_2 \\<br /> \end{array} \right) <br /> <br /> =<br /> \left( \begin{array}{cc}<br /> (k_1 + k_2)/m_1 &amp; -k_2/m_1 \\<br /> -k_2/m_2 &amp; k_2/m_2 \\<br /> \end{array} \right) <br /> <br /> <br /> <br />

So how do we see that the eigenvalues of [m]^{-1}[k] are real and positive?