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 December 15th, 2016, 12:19 AM #1 Newbie   Joined: Dec 2016 From: Haifa Posts: 1 Thanks: 0 Matrix and vectors 1. In given A in size of mXn while m>n and: (A^T)*A=I. What can i say about the eigenvalues of A*(A^T)? and why?
 December 15th, 2016, 06:24 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,276 Thanks: 2437 Math Focus: Mainly analysis and algebra Try writing out an explicit example.
 December 20th, 2016, 12:42 PM #3 Member   Joined: Dec 2016 From: - Posts: 62 Thanks: 10 If your matrix is m x n dimensional, unless m=n I dont see how do you want to get the eigenvalues.... They are only possible to calculate in square matrices
December 20th, 2016, 01:41 PM   #4
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 Originally Posted by nietzsche If your matrix is m x n dimensional, unless m=n I dont see how do you want to get the eigenvalues.... They are only possible to calculate in square matrices
they want the eigenvalues of $A A^T$ which will be $m\times m$

December 20th, 2016, 01:53 PM   #5
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 Originally Posted by romsek they want the eigenvalues of $A A^T$ which will be $m\times m$
you're given $A^T A=I$

you want the eigenvalues of $A A^T$

$A A^T x = \lambda x$

$A^T A A^T x = A^T \lambda x = \lambda A^T x$

$I A^T x = A^T x = \lambda A^T x$

what does this say about $\lambda$ and/or $A^T x$ ?

 December 20th, 2016, 02:14 PM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,276 Thanks: 2437 Math Focus: Mainly analysis and algebra I would just form $A^TA=I \implies AA^TA=AI=A$ so the columns of $A$ are eigenvectors of $(AA^T)$ with eigenvalues of... Thanks from romsek
December 20th, 2016, 02:47 PM   #7
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 Originally Posted by v8archie I would just form $A^TA=I \implies AA^TA=AI=A$ so the columns of $A$ are eigenvectors of $(AA^T)$ with eigenvalues of...
so clever

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