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I am having a matrix A having size n*m, where n>m. I construct two matrices AA' and A'A where A' represents transpose of matrix A. now i find eigen values and eigen vectors for both above created matrix. AA' will have n eigen values and eigen vectors while A'A will have m eigen vectors .i also know that the m eigen values of A'A will also be the eigen values of AA' but i want to know what about the remaining n-m eigen values of AA'. i read that they will zero. is it correct ?if yes, can any explain it simply. I have to study it for use in principal component analysis as it helps in reducing computation.

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More generally, $AB$ and $BA$ will have the same non-zero eigenvalues.

If $AB v = \lambda v$, and $\lambda \neq 0$, then $BABv = \lambda Bv$, and so $\lambda$ is an eigenvalue of $BA$ (since $Bv \neq 0$).

Hence $AA^*$ and $A^*A$ have the same non-zero eigenvalues. Hence all other eigenvalues must be zero.

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  • $\begingroup$ Thank you very much for your answer.i want to ask if there exist any relation between their eigen vectors too? $\endgroup$ – user17815 Mar 3 '14 at 18:02
  • $\begingroup$ Well, if $v$ is an eigenvector (for a non-zero eigenvalue) of $AB$, then $Bv$ will be an eigenvector of $BA$. $\endgroup$ – copper.hat Mar 3 '14 at 18:08
  • $\begingroup$ can we say anything regarding remaining n-m eigen vectors of AA'? $\endgroup$ – user17815 Mar 3 '14 at 18:49
  • $\begingroup$ I'm not sure what you are asking. $\endgroup$ – copper.hat Mar 3 '14 at 18:51

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