I have a a full column rank matrix A, and using this I want to construct a matrix with spectral radius less than 1. I do that using,


H = $I-\alpha A^{T} A$ ($I$ is identity matrix), where the term $\alpha$ is defined as, $0<\alpha < \frac{2}{trace(A^{T} A)}$.


Can I claim that each one of therm in my matrix H will be less than or equal to 1?


I believe that the solution is yes, since I have not been able to come up with the a single scenario where this claim is not true. But I am not able to prove this.



Can we claim that all the terms in a matrix are less than equal to 1 if spectral radius is less than 1?
Yes. Your matrix $H$ is symmetric, so the spectral radius is the same as the maximal eigenvalue is the same as the maximal singular value. From the singular value description, for any unit vectors $\vec{u}$ and $\vec{v}$, we deduce that $|\vec{u}^T H \vec{v}| < 1$. Taking $\vec{u}$ and $\vec{v}$ to be the $i$-th and $j$-th basis vector, we have $|H\_{ij}| < 1$.


