Eigenvalues after setting rows of matrix to zero

Question

I am studying the effect on the eigenvalues of a known Laplacian matrix $L$ when the first $p$ rows (wlog the top rows) are set to $0$. For those not familiar with a Laplacian matrix, all you need to know is that it is a symmetric $(n\times n)$-matrix with ordered spectrum $0=\lambda_1<\lambda_2\leq\cdots\leq\lambda_n$. Note that the eigenvalue $\lambda_2$ is positive - we are working with connected graphs.

I define a projection matrix

$$M=\begin{bmatrix}0_{p\times p} & 0_{p\times(n-p)} \\ 0_{(n-p)\times p} & I_{(n-p)\times(n-p)} \end{bmatrix},$$

which is just the identity matrix with the first $p$ ones set to $0$. Left-multiplication by $M$ has the effect of setting the top $p$ rows of $L$ to $0$, or, equivalently, a projection onto the subspace spanned by basis vectors $p+1$ to $n$.

I postulate that for $p\geq 2$, the matrix $ML$ has eigenvalue $0$ with multiplicity $p$. There are $p$ rows of $0$, so the multiplicity is at least $p$, but I'm stuck on proving equality. Does it have to do with the fact that the rest of the matrix is full rank?

Let me show an example, take

$$L = \begin{bmatrix} 2 & -1 & 0 & 0 & 0 & -1 \\ -1 & 4 & -1 & 0 & -1 & -1 \\ 0 & -1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 & -1 & -1 \\ 0 & -1 & 0 & -1 & 2 & 0 \\ -1 & -1 & 0 & -1 & 0 & 3 \\ \end{bmatrix}$$

which has ordered spectrum $\{0,>0,\cdots, >0\}$. (I don't care what the actual numbers are, just that they are positive/zero)

If $p=1$, I set the first row to $0$, but the spectrum is still $\{0,>0,\cdots, >0\}$. This was a little surprising to me, as I naively expected two $0$'s here, since $0$ was already an eigenvalue, and we in some sense 'added to the nullspace' by setting a row to zero, introducing $0$ as an eigenvalue. What's the intuition behind this?

If $p=2$, I set the first two rows to $0$, and now the spectrum is $\{0, 0, >0, \cdots, >0\}$.

So to summarise:

1. How to prove that $ML$ has eigenvalue $0$ with multiplicity $p$? I understand why it must be $\geq p$.
2. The case $p=1$, what's the intuition for having only multiplicity $1$?

I would appreciate any insights or hints into this problem.

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Related posts here and here.

Answer 1

Let $ML=\pmatrix{0_{p\times p}&0\\ \ast&A}$. Note that for each connected component in the subgraph consisting of the nodes $p+1,\,p+2,\ldots,n$, the corresponding principal submatrix of $A$ cannot be a Laplacian matrix, otherwise the graph for $L$ is not connected. It follows that $A$, up to a relabelling of rows and columns, is a direct sum of irreducibly diagonally dominant matrices. Hence $A$ is invertible and both the algebraic and geometric multiplicity of the zero eigenvalue of $L$ is $p$.