Eigenvalue of ss^T: Why the Maximum Eigenvalue is s^Ts

We can use the spectral theorem to show that the eigenvalues of 'ss^T' are the squared singular values of the matrix 's'. Specifically, if we write the singular value decomposition of 's' as 's = UΣV^T', then we have

$$ss^T = UΣV^TVΣ^TU^T = UΣ^2U^T$$

where 'Σ^2' is the diagonal matrix of squared singular values. Thus, the eigenvalues of 'ss^T' are the diagonal entries of 'Σ^2', which are the squared singular values of 's'.

Since the singular values of 's' are non-negative, we know that the maximum singular value (i.e., the one with the largest magnitude) is equal to ||s||2. Therefore, the largest eigenvalue of 'ss^T' is λ{max}(ss^T) = ||s||_2^2 = s^Ts.