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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 $AA^T$

Mathman2017
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    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. – littleO Jun 02 '17 at 08:50
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    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...., – Chinnapparaj R Jun 02 '17 at 08:51
  • 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. – hardmath Jun 02 '17 at 17:25
  • 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. – hardmath Jun 04 '17 at 15:19

1 Answers1

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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.

User123456789
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