# Largest eigenvalue of matrix equal to 1

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

• It may be helpful to provide the definition of correlation matrix you're using. Jan 24, 2023 at 20:41
• 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).? Jan 24, 2023 at 20:57
• 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.) Jan 24, 2023 at 21:03
• Yes, sorry for unclear notation, square root and quotient are meant elementwise Jan 24, 2023 at 21:07

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 $$S$$.

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