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I have a question like this: We need to use the SVD to prove that if X'X = 0(a matrix with all zeros), where ' means a transpose, then the matrix X = 0; I think may be I need to prove the singular value matrix Sigma will be a matrix with all zeros. Thanks!

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  • $\begingroup$ That's exactly what you need to do. Just remember the first matrix of the decomposition is orthonormal. $\endgroup$ – Koto Sep 14 '17 at 22:07
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    $\begingroup$ Using the SVD is overkill. Assume $X$ is a real $m \times n$ matrix. You can easily show that $X^T X$ and $X$ have the same null space. So if $X^T X = 0$ then the null space of $X^T X$ is $\mathbb R^n$, so the null space of $X$ is $\mathbb R^n$, so $X = 0$. $\endgroup$ – littleO Sep 14 '17 at 22:11
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    $\begingroup$ If $e_i$ is the vector with zeros in all entries but the $i$-th then $Xe_i=x_i$ is the $i$-th column of $X$. Observe that $0=e_i'0e_i=e_i'X'Xe_i=(Xe_i)'(Xe_i)=x_i'x_i=\left\|x_i\right\|^2$. $\endgroup$ – Hellen Sep 14 '17 at 22:12
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$X = U\Sigma V'$.

$0=X'X=V\Sigma^2 V$.

Multiply on the right by $V'$ and on the left by $V'$ to conclude $\Sigma^2=0$, which implies $\Sigma=0$ since it is non-negative.

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  • $\begingroup$ I think the primes are in the wrong places and the sigma should be squared. Otherwise looks good $\endgroup$ – Zach Boyd Sep 14 '17 at 22:22

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