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Given a (real) positive semi-definite matrix $\underline M$ (with all elements of the principal diagonal equal to $1$) and a transformation $T:\underline A\to \underline A + \alpha \underline D$, with $\alpha \in (0,1)$ and $\underline D$ a hollow matrix with all entries one and principal diagonal $0$ (see *), find the biggest $\alpha$ allowed so that $T\underline M$ is again positive semi-definite.

I am not sure if a closed form analytical solution to this problem could even exist. However, here are some ideas that I had in mind.

An upper bound that always ensures PSD:

Let $n=\text{dim}(\underline D)$ and note that $D$ corresponds exactly to the adjacency matrix of the complete graph with $n$-vertices ($1$ denotes a connection and $0$ not). The spectrum of the graph is $\{(-1)^n, (n-1)\}$. Therefore the eigenvalues of $\alpha \underline D$ are $\{ -\alpha,(n-1)\alpha\}$. For PSD we must ensure that all the eigenvalues of $\underline M + \alpha \underline D$ are non-negative, which is the case if the smallest eigenvalue of $\underline M$ is greater than or equal to the "most negative" eigenvalues of $\alpha \underline D$. It then follows:

If $\lambda_{min}\geq \alpha$ then $T\underline M$ is positive semi-definite.

A naïve attempt

Since $T\underline M$ is square, the idea is to use Gershgorin circle theorem to find the intervals that bound the spectrum. Recalling that $M_{ii} = M_{jj} = 1$ for $i\neq j \in 1:n$, we have that

$$\lambda_i\in I\left(1, \sum_{j\neq i}\vert M_{ij} + \alpha\vert\right).$$ Further, we can impose the restriction to have all eigenvalues lie in a positive interval (i.e. the "left endpoint" must be non-negative): $$ \alpha_{max} = \max\bigcap_{i=1}^n \left\{\alpha \in (0,1): \sum_{j\neq i}\vert M_{ij} + \alpha\vert \leq 1\right\}.$$ However, this makes sense only on the specific case that $\underline M$ is diagonally dominant.

Empirically

Running some simulations I have noticed that if all the entries of $\underline M$ are positive, then $\alpha_{max} = \lambda_{\min} + \epsilon$ (for some very small $\epsilon$ - most likely due to some noise). As below

enter image description here

which is not the case if I allow entries to change sign:

enter image description here

in the above plot ($1$ denotes positive semi-definiteness, $0$ not)

(*) For a $3\times 3$ \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{pmatrix}

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  • $\begingroup$ If $M$ is known to be positive semidefinite, then $TM = M + \alpha J$ where $J$ is all-ones matrix, is rank-one perturbation of $M$ with $\alpha\in(0,1)$. Being sum of positive semidefinite matrices, $TM$ should then always be positive semidefinite. $\endgroup$ Nov 10, 2022 at 21:03
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    $\begingroup$ Simulposted to m.se with no notice to either site, math.stackexchange.com/questions/4573715/… $\endgroup$ Nov 10, 2022 at 21:48
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    $\begingroup$ Didn't you just post this question here yesterday? mathoverflow.net/questions/434235/… $\endgroup$ Nov 10, 2022 at 21:51
  • $\begingroup$ Now you know. But why have you posted it to MO twice? $\endgroup$ Nov 11, 2022 at 7:11

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A result assuming that $M$ is positive definite. By continuity, in the optimal $\alpha$ the matrix $M+\alpha D$ is singular; hence the result is the smallest zero of $f(\alpha) = \det (M + \alpha D)$, or, alternatively, of $g(\alpha) = \det(\frac1{\alpha}I + M^{-1/2}DM^{-1/2})$. The latter function vanishes when $\frac{1}{\alpha}$ is an eigenvalue of $-M^{-1/2}DM^{-1/2}$. Hence, $\alpha = \frac{1}{\lambda_{\max}(-M^{-1/2}DM^{-1/2})}$. I don't think it can get any more close than this.

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  • $\begingroup$ 1. Multiply both sides of $f(\alpha)$ by the non-zero quantity $\det M^{1/2}$. 2. Yes, it is the matrix square root (/not/ elementwise). If you prefer, you can take $B$ such that $M = B^T B$ and multiply on the left by $B^{-T}$ and on the right by $B$ instead and the result is unchanged. 3. Yes, $M^{-1/2}$ is a matrix. $\endgroup$ Nov 11, 2022 at 9:52
  • $\begingroup$ Sorry -- multiply both sides by $\det M^{-1/2}$, and divide by $\alpha^n$. No determinant lemmas required. As $\alpha>0$ and $\det M^{1/2}\neq 0$, $g(\alpha)$ vanishes only when $f$ does. $\endgroup$ Nov 11, 2022 at 22:47
  • $\begingroup$ Yes, that works this way, even if that's not clear to me what property you applied to get the last equality. In the version I had in mind, you factor $\frac{1}{\alpha}M+D = M^{1/2}(\frac{1}{\alpha}I + M^{-1/2}DM^{-1/2})M^{1/2}$ and take determinants. $\endgroup$ Nov 11, 2022 at 23:06

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