Similarity Transformation Doesn't seem to work

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Discussion Overview

The discussion revolves around the properties of similarity transformations for square matrices, specifically focusing on the eigenvalues of a matrix A and its similarity transform A' = B-1AB. Participants explore cases where the eigenvalues of A are all zero and question the validity of the transformation in such scenarios.

Discussion Character

  • Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant notes that while similarity transformations should preserve eigenvalues, they encounter an issue when A has all zero eigenvalues, using a specific matrix as an example.
  • Another participant questions the choice of the matrix B used in the transformation and its effect on the eigenvalues of A'.
  • A third participant shares a random matrix T and calculates the eigenvalues of the transformed matrix, revealing small nonzero values, which they suspect may be due to numerical errors.
  • Subsequent comments suggest that the small eigenvalues observed are likely artifacts of numerical inaccuracies rather than true nonzero eigenvalues.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the behavior of eigenvalues under similarity transformations when the original matrix has all zero eigenvalues. There is no consensus on the implications of the numerical results presented.

Contextual Notes

The discussion highlights potential limitations related to numerical precision and the dependence on the choice of the transformation matrix B. The implications of these factors on the eigenvalue calculations remain unresolved.

I_am_learning
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if A is a square matrix, and A' = B-1AB is its similarity transform (with a non-singular similarity transformation matrix B), then the eigenvalues of A and A' are supposed to be the same. I can verify this for all most all cases of A. But, it doesn't seem to work, when the eigen values of A are all zero. For example, it doesn't work when A = [0 1 0;0 0 0; 1 0 0].
Yet I can't see why, it shouldn't work.
from http://mathworld.wolfram.com/SimilarityTransformation.html
the eigen values of A' is given as a solution of
Inline13.gif
=0
Inline13.gif
= 0
Inline15.gif
= 0
Inline18.gif
= 0
Inline21.gif
= 0
Inline24.gif
=0
which is the solution for eigen values of A.
 
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for what matrix B did you find that the eigenvalues of B-1AB to be nonzero?
 
BruceW said:
for what matrix B did you find that the eigenvalues of B-1AB to be nonzero?
I created a random matrix T,
T = rand(3)
T =
0.4509 0.7447 0.1835
0.5470 0.1890 0.3685
0.2963 0.6868 0.6256
and, the eigen values I got are:
eig(inv(T)*A*T)
ans =
1.0e-05 *
0.4418 + 0.0000i
-0.2209 + 0.3826i
-0.2209 - 0.3826i
Oh! now that I post it, I realize that, it may have to do with numerical errors, the values are very small anyway.
Thanks
 
yeah, since the eigenvalues are of order 10^-5, it's likely they are nonzero just due to numerical error.
 

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