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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.

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  • $\begingroup$ If A has eigenspaces of dimension greater than 1 we'd better use an orthonormal basis. $\endgroup$
    – blue
    Aug 27, 2014 at 0:24

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The key point is that row and column sums being zero (equivalent for a symmetric matrix), just means the final eigenvector is $$\frac{1}{\sqrt{n}}\begin{pmatrix} 1 \\ \vdots \\ 1 \end{pmatrix}$$, call this vector $w$ Let $W$ be the $n \times n$ matrix where we have adjoined this eigenvector $w$ as a final column. Your claim is then that in this case

$$ WW^T = I = VV^T + ww^T$$

Which follows just from the rules of matrix multiplication, as pointed out in the comments.

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  • $\begingroup$ At this point what you're claiming is that W is orthogonal because $W W^T = I$ iff $W^T W = I$. But this is obvious since we normalized the eigenvectors and eigenvectors of symmetric matrices with respect to different eigenvalues are orthogonal. $\endgroup$
    – blue
    Aug 27, 2014 at 0:19
  • $\begingroup$ @blue of course, I was invoking the finite dimensional real spectral theorem to get $W$ orthogonal, the real meat of the claim is that $VV^T + ww^T$ is the whole product. $\endgroup$
    – jxnh
    Aug 27, 2014 at 1:57
  • $\begingroup$ @JHance I think you have already proved my claim. Given $WW^T = VV^T + ww^T$, we only need to argue $WW^T=W^TW$ for $W^TW=I$. Since $W$ is orthogonal, W commutes with its transpose, that is $WW^T=W^TW$. $\endgroup$
    – semibruin
    Aug 27, 2014 at 5:13
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    $\begingroup$ @JHance The equation $WW^T = VV^T + ww^T$ follows from elementary matrix multiplication: Every entry of LHS is a dot product of entries of $v_1$ to $v_{n-1}$ plus $1/n$. That the claim is equivalent to $VV^T+ww^T = I$ is also elementary. I don't see any other gaps. $\endgroup$
    – blue
    Aug 28, 2014 at 11:41

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