A confusion related to the rank of the matrix

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Suppose we have a matrix $X$ with $n$ rows and $k$ columns. Further assume that rank of $X$ is $k$ (which means for this particular matrix we must have $n\geq k$). Now the matrix $XX^T$ has $n$ rows and $n$ columns and the rank of $XX^T$ can at most be $k$. So that should mean that $XX^T$ is not invertible (or determinant of $XX^T$ should be zero). Therefore, $XX^T$ can not be a positive semidefinite matrix. But when I read the solution for problem 2.13 of convex optimization book (by Stephen Boyd) I see that they say $XX^T$ is positive definite and its rank is $k$. How is it possible. What is wrong with my thinking here?

Any help in clarifying this confusion will be much appreciated. Thanks in advance.

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Positive semi definite matrices can have zero eigenvalues and be non invertible. Positive definite matrices have only positive eigenvalues and are invertible always. A matrix of the form $M^TM$ is going to be positive semi definite regardless of rank of $M$.