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This question already has an answer here:

How would I go about this proof? I thought it'd be false, but cannot find any counterexamples.

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marked as duplicate by user1551 matrices Sep 26 '17 at 8:37

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  • $\begingroup$ Have you tried using the SVD? $\endgroup$ – Rodrigo de Azevedo Sep 26 '17 at 7:59
  • $\begingroup$ What is the SVD? $\endgroup$ – Kevin Sai Sep 26 '17 at 8:01
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I'll assume that $X$ is a real $m \times n$ matrix. It's very useful to know that $X$ and $X^T X$ have the same null space. Proof: $$ Xu = 0 \implies X^T Xu = 0 \implies u^T X^T X u = 0 \implies \|Xu\|^2 = 0 \implies Xu=0. $$

The fact that you want to prove follows as a corollary. If $X$ has full column rank, then the null space of $X$ is trivial, so the null space of $X^T X$ is trivial, so $X^T X$ is invertible.

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Your intuition is correct. This is indeed false unless further conditions are imposed (such as the matrix is square, or the ground field is real). A simple example is given by the complex matrix $X=\pmatrix{1\\ i}$.

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