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Thanks in advance.

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- Thread starter EvLer
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Thanks in advance.

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JasonRox

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So, the meaning it seems to me is that it maps all it's eigenvectors with respect to the eigenvalue 0 to 0. Therefore, all linear combinations of those eigenvectors are in the kernel of the transformation.

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HallsofIvy

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JasonRox

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It takes the eigenvectors and maps in it to the 0 vector. So, I guess you can say the "collaspe" to the 0 vector. That's probably all the "physical" meaning you can get.

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By definition, an eigenvalue c will be a solution to det(A-cI)=0. If c=0, then det(A)=0.

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JasonRox

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By definition, an eigenvalue c will be a solution to det(A-cI)=0. If c=0, then det(A)=0.

Really?

Let A = cI, then det(A-cI)=0, but det(A) is not equal to 0. How did you deduce that conclusion?

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mjsd

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ok, what I am going to say is not much of a "physical explanation" (it will really depend on the situation and application), but it may help you visualise what is going on.

you start with a set of axes (say the usual x, y), now you can represent an arbitrary vector, v, in this space (defined by your set of axes) using just a set of coordinates (with respect to those set of axes)... eg. v = (a,b).... etc...

now suppose you have an "operator" in this space, represented in a form of a 2x2 matrix. ok, this "operator" can be a representation of anything really (eg. evolution of prey-predator, interactions of system of particles....), it doesn't matter for our discussion here. But what is important is that this "operator" when acted on the vector, v, it

Ok, now, if you look at your eigenvalues equation:

Mu = ku

where u is the eigenvector, k is the eigenvalue, you can see that u is a very special vector in the sense that the "operator" M does nothing but stretch or shrink the length of the eigenvector u by a factor of k (the e-vals)! The moral of this is that if you now express your original vector, v, with respect to the set of axes defined by e-vec u's (instead of the original set with x,y), your entries (a1, b1) where a1, b1 are different from a, b, would be transformed to (k1 a1, k2 b1) by the "operator", where k1 and k2 are the e-vals.

in most instances, such "change of basis" would make the physical interpretation of a system more transparent and simplier to anaylse.

for example, if your M is the moment of inertia matrix/tensor, then diagonalising it give you the principle axes of rotation (the direction of the e-vec's), while the e-vals are the "moment" about those axes.

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Really?

Let A = cI, then det(A-cI)=0, but det(A) is not equal to 0. How did you deduce that conclusion?

What does that have to do with anything? You might want to reread what I said. In particular, what was the constraint on c?

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HallsofIvy

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By definition, an eigenvalue c will be a solution to det(A-cI)=0. If c=0, then det(A)=0.

Really?

Let A = cI, then det(A-cI)=0, but det(A) is not equal to 0. How did you deduce that conclusion?

?? If det(A) is not equal to 0, then c= 0 cannot be an eigenvalue! Ziox

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