For positive semidefinite matrix $L$, it has the property that any $x$, $x^TLx\geq0$. (1)
If L has a eigendecomposition like $LV=V\Lambda$, columns in $V$ are eigenvectors, and $\Lambda$ is a diagonal matrix each element is the eigenvalue. We can easily rewrite the eigen decomposition as $V^TLV=\Lambda$, each element of $\Lambda$: $\lambda_i=v_i^TLv_i$, because $L$ is positive semidefinite, according to (1) $\lambda_i=v_i^TLv_i\geq0$
Because the sum of all rows of the L matrix is 0 (adjacent vertex number is equal to degree), thus the rows are linearly dependent, thus it is not full rank. So there should be at least one eigenvalue be zero.