Gauss-Newton method -- is this matrix product invertible?

Question

In the Gauss-Newton method for solving overdetermined systems of equations, the iteration matrix is of the form $(J^t J)^{-1} J^t$, for a $m \times n$ Jacobian matrix $J$ with $m > n$. I was under the impression that if J had full column rank, then the product $J^t J$ would always be invertible. Is this incorrect?

Answer 1

Since $$\langle J^t Jx, x\rangle = \langle Jx, Jx\rangle = \|Jx\|^2$$ it follows that $\ker J^tJ = \ker J$. In particular, if $\ker J=\{0\}$ ($J$ has full column rank), then $J^tJ$ is invertible.