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Let $S$ be a correlation matrix with positive entries, show that the largest eigenvalue of $\text{diag}\left(\frac{1}{\sqrt{\boldsymbol{S}\boldsymbol{1}}}\right)\boldsymbol{S}\,\text{diag}\left(\frac{1}{\sqrt{\boldsymbol{S}\boldsymbol{1}}}\right)$ is equal to $1$ where diag denotes making a diagonal matrix out of a vector and $\boldsymbol{1}$ denotes the vector which has only 1 entries.

I thought about applying Perron-Frobenius theorem, but it did not help. This Proof that the largest eigenvalue of a stochastic matrix is $1$ seemed also related at the first sight but did not help

Edit: By correlation matrix I mean that it has $1$ entries on the diagonal, is symmetric and positive definite. Squareroot and quotient are meant elementwise

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  • $\begingroup$ It may be helpful to provide the definition of correlation matrix you're using. $\endgroup$ Jan 24, 2023 at 20:41
  • $\begingroup$ I don't understand what the notation means. $S\mathbf{1}$ is a vector, what is its square root (never mind one over the square root).? $\endgroup$
    – Igor Rivin
    Jan 24, 2023 at 20:57
  • $\begingroup$ I think what's intended is $1/\sqrt{\operatorname{diag}(S\mathbf{1})}$, with the properties of correlation matrices ensuring that this diagonal matrix is nonnegative. (That is, the root and reciprocal are intended as element-wise.) $\endgroup$ Jan 24, 2023 at 21:03
  • $\begingroup$ Yes, sorry for unclear notation, square root and quotient are meant elementwise $\endgroup$
    – Philipp123
    Jan 24, 2023 at 21:07

1 Answer 1

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So, let

$$P=\mathrm{diag}_i\left(\dfrac{1}{\sqrt{\sigma_i}}\right)$$

where $\sigma_i=\sum_{j=1}^ns_{ij}$.

We need to check the maximum eigenvalue of $PSP$. The eigenvalues are the same as

$$ SP^2=S\mathrm{diag}_i\left(\dfrac{1}{\sigma_i}\right). $$

Now we have that

$$ S\mathrm{diag}_i\left(\dfrac{1}{\sigma_i}\right)\sigma=S\mathrm{1}=\sigma. $$

Since $SP^2$ is a positive matrix, or at least nonnegative, and we have that at all diagonal entries are equal to 1, then $\sigma$ is a positive vector and, by virtue of the Perron-Frobenius theorem, this implies that 1 is the dominant eigenvalue of $S$.

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  • $\begingroup$ Thanks for the answer, I cannot see how the eigenvalues of $PSP$ are the same as the values of $SP^2$ $\endgroup$
    – Philipp123
    Jan 24, 2023 at 21:31
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    $\begingroup$ Similarity transformation. $\endgroup$
    – KBS
    Jan 24, 2023 at 22:17

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