$I$ is an identity matrix of size $n \times n$. $X$ is a matrix of size $n \times k$(Assuming $k \leq n$).
As we know, $(I+XX')$ is invertible. Because $(I+XX') = (I(blank)X)*(I(blank)X)'$, where $(I (blank)X)$ has full rank of n.
So I'm wondering whether $(I-XX')$ has a similar property? What's the invertibility of matrix $(I - XX')$?
$I-XX'$ need not be invertible. For example, take $X=I$.
More generally, take a matrix $P$ such that $PP'=I$.
To fully answer your question, note that $XX'$ is a real symmetric matrix, and can be diagonalized. So, $I-XX'$ is not invertible precisely when there exists at least one eigenvalue of $XX'$ equal to $1$.