For a $m \times n$ matrix $A$, what's the relationship of $A*A^T$ and $A^T*A$? The background of this question is that if we see the row of $A$ as observations and column as variables, $A*A^T$ is the covariance across variables, and $A^T*A$ are the covariance across observations. How is these two matrix related?
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@Gabriel, these are not transposes of each other. $(A^T \cdot A)^T = A^T \cdot (A^T)^T = A^T \cdot A$. Also they are of different sizes. – Joel Mar 19 '14 at 14:43
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I don't know what you're looking for, but
- $\text{rank}\left(AA^T\right)=\text{rank}\left(A^TA\right)=\text{rank}\left(A\right)=\text{rank}\left(A^T\right),$
- The matrices $AA^T$ and $A^TA$ share the same non-null eigenvalues (they also share multiplicities of these eigenvalues).
- Since both are symmetric matrices, they both produce an orthonormal basis of eigenvectors of $\mathbb R^m$ and $\mathbb R^n$, respectively for $AA^T$ and $A^TA$. In the notation of the SVD decomposition article on wikipedia ($A=U\Sigma V^T$). The orthonormal eigenvectors of $AA^T$written by columns in a matrix yield $U$, where as those of $A^TA$ yield $V$.
Git Gud
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Git Gud's answer is in a sense all there is to say: given any $m \times m$ and $n \times n$ positive semidefinite matrices $B$ and $C$ with the same nonzero eigenvalues (counted by multiplicity), there is some $m \times n$ matrix $A$ with $A A^T = B$ and $A^T A = C$.
Robert Israel
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only require positive semidefinite, not symmetric? could you please point me a proof? thanks, – ahala Mar 19 '14 at 18:02
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Sorry, I guess I should have specified: I'm using a definition of positive semidefinite that includes being symmetric. Just use unitary diagonalizations $B = U \Sigma_m U^T$ and $C = V \Sigma_n V^T$, and take $A = U D V^T$ where $D$ is $m \times n$ with diagonal elements the square roots of the nonzero eigenvalues (plus as many zeros as needed). – Robert Israel Mar 20 '14 at 01:18