similar to an eigenvalue problem . . . help!

**BobbyBear** · Nov28-09, 03:59 PM

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) :

$LaTeX Code: \\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:

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

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

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

where

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

and

$LaTeX Code: \\vec{\\phi _n} $

is a constant vector.

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

Substituting into the differential system,

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

from which

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

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

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

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

from which we get two values of

$LaTeX Code: \\omega _n^2 $

Now, my book (Dynamics of Structures by Chopra) says that the $LaTeX Code: \\omega _n^2$ 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 $LaTeX Code: \\omega _n^2$ 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** · Nov29-09, 10:27 PM

The matrix

is invertible. Factor it out of

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

to get

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

or

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

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

**BobbyBear** · Nov30-09, 06:55 AM

Thank you Mute, I never thought of doing that!

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

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

So we have that both $LaTeX Code: [m]^{-1} $ and

are positive definite, but in general that does not mean that $LaTeX Code: [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

, but this is not the case with $LaTeX Code: [m]^{-1} $ and [k], their product is not commutative in general. So how can we see that $LaTeX Code: [m]^{-1} [k] $ is positive definite?

Thanks for your help!

**BobbyBear** · Nov30-09, 07:12 AM

Oh but wait! I just realised that even though [m] and [k] are real symmetric, $LaTeX Code: [m]^{-1}[k]$ is not even symmetric, so it wouldn't be of any use to prove that $LaTeX Code: [m]^{-1}[k]$ is positive definite, would it, because we don't know that the eigenvalues are real...

$LaTeX Code: [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 $LaTeX Code: [m]^{-1}[k]$ are real and positive?