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What does the Eigenvalue of a linear system actually tell you?

by newclearwintr
Tags: eigenvalue, linear
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newclearwintr
#1
Jan23-13, 04:50 PM
P: 3
I know that the eigenvalue of a linear system is a scalar such that Ax=λx. I know many ways to find the eigenvalue of a linear system. But I'm pulling my hair out trying to figure out what it is actually telling me about the system.

Can anyone give me a non-technical straight up answer on why eigenvalues are important and what that information provides us in regards to a linear system?

Thanks!
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mathwonk
#2
Jan23-13, 05:14 PM
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well if you had n independent eigenvectors all with eigenvalue 1, then A = Id.

If you have an orthogonal matrix in dim n= 2, and one eigenvalue is 1 and one eigenvalue is -1, you have a reflection.

...... it helps describe the geometry of the map.
tiny-tim
#3
Jan23-13, 05:17 PM
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hi newclearwintr! welcome to pf!

if A is a force or torque, then the eigenvalue is the effective mass or effective moment of inertia in a particular mode

if A is a derivative, then the eigenvalue is the time constant in a particular mode

(the only modes that will work are the eigenvectors

if the system starts in any other mode, it won't stay in it, so the concept of effective mass or whatever is inapplicable)

newclearwintr
#4
Jan23-13, 06:09 PM
P: 3
What does the Eigenvalue of a linear system actually tell you?

Quote Quote by tiny-tim View Post
hi newclearwintr! welcome to pf!

if A is a force or torque, then the eigenvalue is the effective mass or effective moment of inertia in a particular mode

if A is a derivative, then the eigenvalue is the time constant in a particular mode

(the only modes that will work are the eigenvectors

if the system starts in any other mode, it won't stay in it, so the concept of effective mass or whatever is inapplicable)
Thanks for your response tiny-tim! So when you say it won't stay in the mode, we are saying that the mathematical model used to describe whatever phenomenon we're talking about will no longer be applicable? That constant is what allows the system to work (ie stay in the mode)?
tiny-tim
#5
Jan23-13, 06:35 PM
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hi newclearwintr!

no, the mathematical model is fine

suppose the equation is L = Iω (angular momentum = moment of inertia times angular velocity)

the I here is a tensor, not a number, and L is generally not parallel to ω

only if ω is parallel to a principal axis of the body, is L parallel to ω

ie the only modes with L parallel to ω are rotations about the principal axes (the eigenvectors)

for any other mode, even if angular momentum is constant (conserved), the angular velocity won't be the rotation won't stay in the mode it started in!
Ferramentarius
#6
Feb4-13, 02:01 AM
P: 22
The trace of matrix A which equals the sum of its diagonal elements equals the sum of all eigenvalues of the matrix and the determinant of A equals the product of the eigenvalues. A is invertible if and only if all the eigenvalues are nonzero. The solutions of the characteristic polynomial of A are exactly its eigenvalues while a real matrix B is diagonalizable if all its eigenvalues are real and distinct. A symmetric matrix can be diagonalized in case its eigenvalues are distinct, eigenvalues of Hermitian matrices are real, and eigenvalues of skew-symmetric matrices are purely imaginary or zero. When A is invertible the eigenvalues of A^-1 are precisely the multiplicative inverses of the eigenvalues of A.

Planetmath: eigenvalues (of a matrix)
Wolfram Mathworld: Eigenvalue
Wikipedia: Eigenvalues
Robert1986
#7
Feb4-13, 06:24 AM
P: 828
I guess one way to think about eigenvectors is that they are vectors that the matrix only stretches or contracts. If the eigenvalue associated to an eigenvector is less than one, the matrix contracts the vector (makes it shorter) and if the eigenvalue is bigger than 1, the matrix stretches the vector. Of course, if the eigenvalue is 1, the matrix is just the identity on that vector.


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