Eigenvalue in 3D

  • Thread starter gaffareee
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  • #1
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Dear frnds,

suppose one have a 3D matrix,

A=ones(3,3,3);

he wants to have eigen value of A,
then,

A-Lamda*I=0 where A is 3D matrix, I is 3D matrix.
now

Problem is our mathematical soft can only do 2D inversion or eigen calculation.
Please refer me any info to find out the eigenvalue of 3d matrix.

thanks anyway.
 

Answers and Replies

  • #2
HallsofIvy
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I'm sorry, what do you mean by "ones(3,3,3)"? if there were only two "3"s I would interpret that as
[tex]\begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1\\ 1 & 1 & 1\end{bmatrix}[/tex]
the 3 by 3 matrix consisting of all "1"s.

Do you only know how to use software to find eigenvalues? What about determinants? I think you have found a problem with relying on software! Do you know the definition of "eigenvalue"?

One of the things you need to know is that [itex]A- \lambda I[/itex] does NOT have to be the 0 vector, although, if you really meant "ones(3,3)", that does happen. A number, [itex]\lambda[/itex], is an eigenvalue if and only if there exist a non-zero vector v such that [itex](A- \lambda I)v= 0[/itex].

To find the eigenvalues of a 3 by 3 matrix, use the fact that if [itex](A- \lambda)v= 0[/itex] for a non-zero vector, then the matrix [itex]A- \lambda I[/itex], which is just A with the variable [itex]\lambda[/itex] subtracted from each of its diagonal elements), while it is not necessarily 0, cannot have an inverse and so its determinant must be 0.

For the matrix above, that means you need to solve
[tex]\left|\begin{array}{ccc}1-\lambda & 1 & 1 \\ 1 & 1-\lambda & 1 \\ 1 & 1 & 1-\lambda\end{array}\right|= 0[/tex]
That will give a cubic equation in [itex]\lambda[/itex] which might have up to three different roots. Since it should be clear that the determinant of A itself if 0, one of those roots should be obvious.
 
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  • #3
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thanks for your kind reply.
 

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