I would like to show that there exists some scalar $\eta>0$ such that $\eta A^TA \preccurlyeq \frac{1}{2}(A+A^T)$ for any square matrix $A$ with nonnegative eigenvalues that are bounded by 1 $(0\leq \lambda(A)\leq 1)$. $A$ is not necessarily symmetric. If it helps $A=I-W$ where $W$ is a row stochastic matrix with positive entries.
I tried to rely on the following: for any two PSD matrices $U,V$ it holds that $\sigma(UV)\leq \frac{1}{2}\sigma(U^2+V^2)$ where $\sigma(C)$ represents the singular value of the matrix $C$. However, $A$ is not symmetric in general.
I would appreciate any help. Thank you!
This is not always possible, as $\eta A^TA$ is always positive semidefinite but $\frac12(A+A^T)$ isn't, such as when $$ A=I-\pmatrix{1&0\\ 1&0}=\pmatrix{0&0\\ -1&1}. $$ By perturbing the $W$ above slightly, you may also obtain a counterexample in which $W$ is entrywise positive, i.e. you may consider $$ A=I-\pmatrix{1-\epsilon&\epsilon\\ 1-\epsilon&\epsilon}=\pmatrix{\epsilon&-\epsilon\\ \epsilon-1&1-\epsilon} $$ for some sufficiently small $\epsilon>0$.