Can we claim that all the terms in a matrix are less than equal to 1 if spectral radius is less than 1?

Question

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, $\alpha = \frac{2}{trace(A^{T} A)}$.

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.

Answer 1

The answer is yes.

Claim: $\|B\| \leq 1 \implies |B_{ij}| \leq 1$.

Proof: Note that $$ |B_{ij}| = |e_i^TBe_j| = \|e_i^TBe_j\| \leq \|e_i^T\| \, \|Be_j\| \leq \|e_i^T\|\,\|B\|\, \|e_j\| \leq (1)(1)(1) = 1 $$

Answer 2

Let

$$A= \left(\begin{array}{cc} \frac 12 & 100 \\ 0 & \frac 12\end{array}\right),$$

It is upper diagonal, so eigenvalues are the diagonal elements. Spectral radius is $\frac 12$.