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If a matrix $A$ has linearly independent columns, why is $A^TA$ invertible?

I think it has to do with $A^TA$ being square and that $A^TAx$=0 only has the trivial solution, but I don't really know how to explain that properly.

user3427042
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1 Answers1

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$A^TA$ is not always invertible even if $A$ has full column rank. Counterexample: $A=\pmatrix{1\\ i}$ over $\mathbb C$ or $A=\pmatrix{1\\ 1}$ over $GF(2)$.

Over the reals, $A^TA$ is invertible if $A$ has full column rank. Hint: observe that $\|Ax\|^2=x^TA^TAx$ where $\|\cdot\|$ is the Euclidean norm.

user1551
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