In singular value decomposition (SVD), $X=USV^T$, if $X$ is $N\times D$, then $U$ is $N\times N$, $S$ is $N\times D$, and $V$ is $D\times D$.
Let's assume $N<D$.
Every column vector in $V$ is an eigenvector of $X^TX$, so we have $D$ eigenvectors in total.
Each (squared) diagonal element of $S$ is an eigenvalue of $X^TX$, so we have $\min(N, D)=N$ eigenvalues.
Here comes my question: why do we have unequal numbers of eigenvalues and eigenvectors? Shouldn't each eigenvector be associated with one eigenvalue?