Let $A$ be an $n \times n$ real symmetric matrix with row and column sums zero. For example, $$ A=\begin{bmatrix}1 & -2 & 1\\ -2 & 1 & 1\\ 1 & 1 & -2 \end{bmatrix}. $$ I have the following interesting observation about $A$ in general.
Claim: Suppose $\mathrm{rank}(A)=n-1$, and let $v_1, v_2,\dots ,v_{n-1}$ be the $n-1$ normalized eigenvectors (with unit length) corresponding to the $n-1$ nonzero eigenvalues. Let $\mathbf{V}=[v_1,\dots,v_{n-1}]$ be an $n\times(n-1)$ matrix of which each column $i$ is the eigenvector $v_i$. We have $$ I-\mathbf{V}\mathbf{V}^{T}=\frac{1}{n}\begin{bmatrix}1 & \dots & 1\\ \vdots & \vdots & \vdots\\ 1 & \dots & 1 \end{bmatrix}, $$ where $I$ is the identity matrix, and $n$ is the number of columns.
As for our particular $A$ in the display, we have $$ \mathbf{V}=\begin{bmatrix}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}}\\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}}\\ 0 & -\frac{2}{\sqrt{6}} \end{bmatrix}. $$ One can easily verify the above claim for this example. I have randomly generated many such matrices, and the claim holds. So it might be correct.
My question is how to prove it. After spending many hours, I have made little progress so far. The only thing meaningful I have found is that any eigenvector of $A$ must sum to be zero, because $0=1^TAv=\lambda 1^Tv$. Here $\{\lambda, v\}$ denotes a generic pair of eigenvalue and eigenvector. An additional observation is that all cofactors of $A$ are identical. But these observations are far from enough to understand this claim. Any thought is welcomed. Thanks.
