If $X$ is a $n$ by $k$ real matrix, and we know that $X^{T}X$ is invertible, is $X$ invertible as well? If $X$ is a $n$ by $k$ real matrix, and we know that $X^{T}X$ is invertible, we know that $rank(X^{T}X) = n$ by $n$. Can we say that $X$ is invertible as well and hence has rank $n$? thanks!
$$\bf X=\begin{bmatrix}1\\0\end{bmatrix}\,:\mathbb R\to\mathbb R^2\\{\bf X}^T\bf X=I:\mathbb R\to\mathbb R $$
But X is not surjective so has no right inverse.
No, if I understand the question right, it is not possible to say that $\mathbf{X} $ is invertible and also $\mathbf{X}^T\mathbf{X}$ is only invertible when $\mathbf{X}$ has full column rank. In that case, you can find the Moore Penrose pseudo inverse of $\mathbf{X}$.