Could you please help me to prove or disprove this proposition?

1. Let $A$ be an $n×n$ matrix. Then the eigenvalue(s) of $A^TA$ is the same as the eigenvalue(s) of $AA^T$


closed as off-topic by Surb, kingW3, Arnaldo, B. Mehta, hardmath Jun 2 '17 at 17:26

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Arnaldo, B. Mehta, hardmath
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ You're more likely to get a good response if you describe your attempts and your thoughts about the problems, and what specifically you got stuck on. $\endgroup$ – littleO Jun 2 '17 at 8:50
  • 1
    $\begingroup$ The non-zero eigenvalues of $AA^T$ and $A^TA$ are necessarily the same..Start with $\lambda \neq 0$ be an eigenvalue of $AA^T$. Then $AA^T x=\lambda x$. Multiplying both sides by $A^T$, we have......... and conclude that...., $\endgroup$ – Chinnapparaj R Jun 2 '17 at 8:51
  • $\begingroup$ A good rule of thumb is not to pile on multiple problems in one Question unless you've done enough analysis to know that they have a substantial relationship. I'm not seeing that to be the case here. $\endgroup$ – hardmath Jun 2 '17 at 17:25
  • $\begingroup$ Related: What is the difference between singular value and eigenvalue? Note that the eigenvalues of $A^TA$ are precisely the squares of the singular values of $A$. When $A$ is a square matrix, the singular value decomposition (SVD) of $A$ makes it evident that $A^T$ shares the same singular values. The desired result follows easily. $\endgroup$ – hardmath Jun 4 '17 at 15:19

Some hints:

For 1., you should rewrite the matrices $A$ and $A^T$ in their diagonalised form. Compute the products of $AA^T$ and $A^TA$. What do you notice?

For 2., the characteristic polynomial of a $5\times 5$ matrix will be fifth order. What do you know of complex roots and their conjugates?

For 3., note that $Av=\lambda v$ if $\lambda$ is an eigenvalue. What do you know about the definition of the null space and range? Can you make the connection with the eigenvector product?

For 4., I myself am not familiar with the notation $A$~$B$. Like littleO mentioned, try to show your own efforts, this would make the community more responsive, or I would have been able to give a hint on how to approach the problem.

  • $\begingroup$ ~ denotes similarity of two matrices $\endgroup$ – Mathman2017 Jun 2 '17 at 9:08
  • $\begingroup$ In that case, note that similar matrices have a lot of common properties. Connect this to the other propositions. $\endgroup$ – User123456789 Jun 2 '17 at 9:23
  • $\begingroup$ Note that 1. is true even if matrix $A$ (equiv. matrix $A^T$) is not diagonalizable. $\endgroup$ – hardmath Jun 3 '17 at 4:45
  • $\begingroup$ Since the Question remains closed, I've posted a Comment above, outlining the proof of the generally true claim. $\endgroup$ – hardmath Jun 4 '17 at 15:29

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