Assume we have a matrix $A$ let's say $100 \times 4$. We determine the product $B=A^{T}A$ Then by the spectral theorem
\begin{equation} B =U^{T} \lambda U \end{equation}
$B$ is a symmetric matrix $4 \times 4$. Is it impossible that is not invertible?
I was thinking that we cannot guarantee that is always invertible because maybe one of the eigenvalues is zero. But I am not sure if this is correct. In case one of the eigenvalues is zero then we need to calculate the pseudo-inverse?
Any help or comment will be helpful.
It is possible that $B$ is non-invertible. Suppose every entry of $A$ was the same value, then every entry of $B$ is the same value, and so it is not invertible.
More generally, if $A$ does not have full column rank (4), then $B$ will not be full rank, and thus not be invertible.