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I need some help on showing that $A^TA$ has at least one positive eigenvalue if $A$ is not all-zero. $A$ is rectangular and can have dependent columns in general. I can show that it cannot have negative eigenvalues. Here is what I have now.

$$ A^TA\vec x=\lambda \vec x\\ \vec x^TA^TA\vec x=\vec x^T\lambda \vec x\\ ||A\vec x||^2=\lambda||\vec x||^2 $$ Because the two norms are strictly non-negative, $\lambda$ has to be non-negative. However, I cannot guarantee that $||A\vec x||\neq 0$ for at least one $\vec x$. If all eigenvectors of $A^TA$ lies in the null-space of $A$ (which I feel should not be the case), then $\lambda$ is indeed zero for all eigenvectors. So is my feeling wrong or how do I prove it?

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  • $\begingroup$ Since $A$ isn't the null matrix, there are $i,j$ such that the $(i,j)$-entry, $a_{ij}$, isn't null. Let $x=\vec e_j$. $\endgroup$
    – Git Gud
    Commented Mar 10, 2015 at 18:22
  • $\begingroup$ @GitGud Then is $\vec x=\vec e_j$ an eigenvector of $A^TA$? I don't think so... $\endgroup$
    – zyl1024
    Commented Mar 10, 2015 at 18:30
  • $\begingroup$ No, then $A\vec x$ isn't null. ("However, I cannot guarantee that $||A\vec x||\neq 0$ for at least one $\vec x$") $\endgroup$
    – Git Gud
    Commented Mar 10, 2015 at 18:33
  • $\begingroup$ I feel a bit lost on the logic. So from the beginning we let $\vec x$ be an eigenvector for $A^TA$. For any non-null $A$ you will find a $\vec x$ such that $||A\vec x||\neq 0$. But don't we need to also guarantee that the $\vec x$ we find is an eigenvector of $A^TA$? $\endgroup$
    – zyl1024
    Commented Mar 10, 2015 at 18:37
  • $\begingroup$ As I see it you proved that given an eigenvalue $\lambda$, $\forall\vec x(A^TA\vec x=\lambda \vec x\implies \Vert A\vec x\Vert ^2=\lambda \Vert \vec x\Vert ^2)$.You seem to be incorrectly using the direction $\Longleftarrow$. $\endgroup$
    – Git Gud
    Commented Mar 10, 2015 at 18:40

1 Answer 1

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Note that $A^TA$ is symmetric and positive semidefinite: for any conforming $v$, it follows that $v^TA^TAv=|Av|^2\geq0$. Thus, the eigenvalues of $A^TA$ are all nonnegative. (We can actually prove this directly by looking at $A^TAv=\lambda v\implies0\leq|Av|^2=\lambda|v|^2$).

So if $A^TA$ doesn't have a positive eigenvalue, then its eigenvalues are all $0$. By the diagonalisation of symmetric matrices, we can infer that $A^TA$ is $0$. But if we inspect the diagonal elements of $A^TA$ and use the matrix multiplication formula, it becomes evident that $A$ itself must be $0$. This is a contradiction, so the claim follows.

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