Using SVD to prove that a matrix X will be all zeros if X'X is a matrix with all zeros

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!

• That's exactly what you need to do. Just remember the first matrix of the decomposition is orthonormal. – Koto Sep 14 '17 at 22:07
• 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 $X = 0$. – littleO Sep 14 '17 at 22:11
• 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 $0=e_i'0e_i=e_i'X'Xe_i=(Xe_i)'(Xe_i)=x_i'x_i=\left\|x_i\right\|^2$. – Hellen Sep 14 '17 at 22:12

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