Does the Matrix xyTA Have More Than One Non-Zero Eigenvalue?

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Discussion Overview

The discussion centers around the properties of the matrix product xyTA, where x and y are vectors and A is a square matrix. Participants explore whether this matrix can have more than one non-zero eigenvalue, examining concepts related to matrix rank and eigenvalues.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant asserts that the matrix xyT has a rank of 1, leading to the conclusion that the rank of xyTA is also at most 1, which implies at most one non-zero eigenvalue.
  • Another participant questions the interpretation of xyT, suggesting that it yields a scalar, which could imply a different understanding of eigenvalue count.
  • A later reply clarifies that xTy is indeed a scalar, but does not resolve the implications for the eigenvalues of xyTA.

Areas of Agreement / Disagreement

Participants express differing views on the implications of the rank of xyT and its relationship to the eigenvalues of xyTA. The discussion remains unresolved regarding the maximum number of non-zero eigenvalues.

Contextual Notes

There are assumptions regarding the definitions of rank and eigenvalues that are not fully explored, and the implications of scalar products on eigenvalue counts are not clearly established.

Leo321
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We have vectors x,y of size n and a matrix A of size nxn.
Is it true that the matrix xyTA has at most one non zero eigenvalue? Why is it so?
 
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Ok, I think I got it.
The matrix xyT has a rank of 1.
It is known that rank(AB)<=min(rank(A),rank(B))
Thus rank(xyTA)<=1
It is also known that the number of non-zero eigenvalues of a matrix is less or equal to the matrix rank. Thus the number of non-zero eigenvalues of xyTA is at most 1.
Right?
 
xy(transpose) will yield a scalar correct?

That means the maximum number of eigenvalues is n i believe
 
khemist said:
xy(transpose) will yield a scalar correct?


no, xTy is a scalar
 

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