I am trying to solve the following problem.
Let the following incidence matrix of an undirected graph with four nodes $$ B = \begin{bmatrix}1 & 0 & -1 & 0 & 0 \\ -1 & 1 & 0 & 1 & 0 \\ 0 & -1 & 1 & 0 & 1 \\ 0 & 0 & 0 & -1 & -1 \end{bmatrix}, $$ and now I set all the elements of the first row of $B$ to zero, therefore we have the following "incidence" matrix $$ B_d = \begin{bmatrix}0 & 0 & 0 & 0 & 0 \\ -1 & 1 & 0 & 1 & 0 \\ 0 & -1 & 1 & 0 & 1 \\ 0 & 0 & 0 & -1 & -1 \end{bmatrix}. $$
It is straightforward to check that $BB_d^T$ looks like the Laplacian matrix $L = BB^T$ but with all the elements of first column set to zero. Also note that $B^TB_d$ is symmetric.
My question is how to prove that $B^TB_d$ is positive semidefinite?
We know that the non-zero eigenvalues of $BB_d^T$, which are positive just applying Gershgorin circle theorem to its columns (easy to check since all the columns excepting the first one are equal to L) are also present in $B^TB_d$. The problem is that $B^TB_d$ is bigger than $BB_d^T$, so I am missing the other two eigenvalues.
Numerically it can be shown that the eigenvalues of $B^TB_d$ are the same as $BB_d^T$ plus one zero, so it is the spectrum of $L$ from the 'directed graph' and an additional zero.