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I know that $A^TA$ is positive semidefinite matrix, however I was wondering if this reasoning is incorrect since $A,A^T$ may have different eigenvectors.

Just saying $A^TAx=\lambda^2x$ where $\lambda$ is an eigenvalue of $A,A^T$ and $x$ a corresponding eigenvector. The only reason I thought this would be acceptable is because I know when considering the quadratic form of $A$, we can assume the matrix is symmetric. Is it true the product of two matrices being symmetric implies both are symmetric? So why would my assumption be incorrect here?

Scott Frazier
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  • Check this: https://math.stackexchange.com/q/730421/42969, or this: https://math.stackexchange.com/q/133350/42969, or this: https://math.stackexchange.com/q/812624/42969 – Martin R Mar 17 '21 at 21:23
  • @MartinR I can see the correct way to do it from that, but still cannot decide whether my approach is incorrect and why. – Scott Frazier Mar 17 '21 at 21:36
  • "Is it true the product of two matrices being symmetric implies both are symmetric?" No, that's not true. In your context, $A$ need not even be square. Look at a simple example: $A = \begin{pmatrix} 1 & 2 \end{pmatrix}$ with $A^TA = \begin{pmatrix} 1 & 2 \ 2 & 4 \end{pmatrix}$. – cavok Mar 17 '21 at 21:40
  • @cavok so lets get to the point the explanation is wrong. That's all I am wondering – Scott Frazier Mar 17 '21 at 21:41
  • Well, if $A$ is square and symmetric, there's not much to show to start with, as the eigenvalues of $A^TA = A^2$ are the squares of the eigenvalues of $A$. If $A$ is square but not symmetric, then as you write, the eigenvectors of $A^T$ are almost always different from those of $A$, so there's no reason that eigenvectors $x$ of $A$ should satisfy $A^TA x = \lambda^2 x$. – cavok Mar 17 '21 at 21:48

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