$X$ is an $n \times m$ matrix, where $n \geq m$ and $\mbox{rank}(X) = m$.
Is it possible for $X^TX$ to not be full rank?
If $X$ can only be square I could easily prove this with $$ \det(X^TX) = \det(X^T)\det(X) = \det(X)^2 > 0 $$
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If you are talking about a real matrix $X$, no.
If $X^TXv=0$, then $\|Xv\|_2^2=v^TX^TXv=0$ and hence $Xv=0$. Since $X$ has full column rank, $v$ must be zero. Thus $X^TX$ is nonsingular and it has full rank.
user1551
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Wait, how does this show $X^TX$ is invertible? Is it because you showed that $v$ must be 0? – 24n8 Aug 04 '20 at 20:16
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@Iamanon The answer shows that $X^TXv=0\Rightarrow v=0$, meaning that $X^TX$ is nonsingular. Over a field, a square matrix is nonsingular if and only if it is invertible. – user1551 Aug 04 '20 at 20:20
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No, it is not possible as long as $X$ is a real matrix. Because, we have: $$rank(X^\top X) = rank(X).$$ Hence, $rank(X^\top X) = m$ too. Since $X^\top X$ is $m\times m$, it must be of full rank.
How to show that $rank(X^\top X) = rank(X)?$ Show that their null spaces are same. Of course, $null(X) \subseteq null(X^\top X)$. For the reverse direction, note that if $X^\top X v = 0$, then $v^\top X^\top X v = 0$, and hence, $\|Xv\|^2 = 0$, hence $Xv = 0$ (here you use the fact that the entries of $X$ are real).
Counterexample for complex matrices: Let $X$ be a $2\times 2$ matrix, with $X_{1,1}=i$, $X_{1,2}=X_{2,1} = 1$ and $X_{2,2}=-i$, where $i$ is the root of $-1$. Then, $X$ has rank $1$ (why?) but $X^\top X$ has rank $0$.
Probabilist
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Does the same proof work with $X^\intercal$ replaced by $X^*$ when $X$ has complex entries? – Carl Schildkraut May 09 '20 at 02:28
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1Yes, basically in that case, X has a singular value decomposition as UDV*, and @Siong Thye Goh's proof works with that modification. – Probabilist May 09 '20 at 02:39
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It's been awhile since I've dealt with classical linear algebra concepts. Could you explain why $null(X) \subseteq null(X^TX)$? I remember that the nullspace of a matrix $A$ is all $x$ such that $Ax=0$ and $x \neq 0$ – 24n8 May 09 '20 at 02:50
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