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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.

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.

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.

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

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.