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BobbyBear
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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) :

[tex] \left( \begin{array}{cc}<br /> m_1 & 0 \\<br /> 0 & 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) & -k_2 \\<br /> -k_2 & 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 /> [/tex]

which I shall compactly write as:

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

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

[tex] \vec{u}(t)=q_n(t) \vec{\phi _n} [/tex]

where

[tex] q_n(t) = A_n cos (\omega _n t) + B_n sin (\omega _n t)[/tex]

and

[tex] \vec{\phi _n} [/tex]

is a constant vector.

Then
[tex] \vec{\ddot{u}}(t)=-\omega _n^2 q_n(t) \vec{\phi _n}[/tex]

Substituting into the differential system,

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

from which

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

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

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

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

from which we get two values of

[tex] \omega _n^2[/tex]

Now, my book (Dynamics of Structures by Chopra) says that the [tex]\omega _n^2[/tex] 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 [tex]\omega _n^2[/tex] 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?
 
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The matrix [itex][m][/itex] is invertible. Factor it out of

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

to get

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

or

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

which means the [itex]\omega_n^2[/itex] are eigenvalues of the matrix [m]^{-1}[k].
 
Thank you Mute, I never thought of doing that!

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

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

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 [itex] [m]^{-1}[/itex]

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

Thanks for your help!
 
Oh but wait! I just realized that even though [m] and [k] are real symmetric, [itex][m]^{-1}[k][/itex] is not even symmetric, so it wouldn't be of any use to prove that [itex][m]^{-1}[k][/itex] is positive definite, would it, because we don't know that the eigenvalues are real...

[tex][m]^{-1}[k] =<br /> <br /> \left( \begin{array}{cc}<br /> 1/m_1 & 0 \\<br /> 0 & 1/m_2 \\<br /> \end{array} \right) <br /> <br /> \left( \begin{array}{cc}<br /> (k_1 + k_2) & -k_2 \\<br /> -k_2 & k_2 \\<br /> \end{array} \right) <br /> <br /> =<br /> \left( \begin{array}{cc}<br /> (k_1 + k_2)/m_1 & -k_2/m_1 \\<br /> -k_2/m_2 & k_2/m_2 \\<br /> \end{array} \right) <br /> <br /> <br /> [/tex]

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