# Prove that $X^TX$ is invertible when $X$ has full column rank. [duplicate]

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

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• Have you tried using the SVD? – Rodrigo de Azevedo Sep 26 '17 at 7:59
• What is the SVD? – Kevin Sai Sep 26 '17 at 8:01

## 2 Answers

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.

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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