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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.

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  • $\begingroup$ No I mean how can I prove that the values in H are bounded by 1? $\endgroup$
    – Rohit Shukla
    Commented Mar 29, 2016 at 2:50
  • $\begingroup$ Sorry, I misread the question. I have deleted my comment. $\endgroup$
    – LSpice
    Commented Mar 29, 2016 at 2:51

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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 $|H_{ij}| < 1$.

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    $\begingroup$ To say it another way, for a symmetric matrix, the spectral radius equals the operator norm. And by Cauchy-Schwarz $|u^T H v| \le |u| |Hv| \le |u| |v| \|H\|$. Then take $u,v$ to be unit basis vectors. $\endgroup$
    – Nate Eldredge
    Commented Mar 29, 2016 at 4:10
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    $\begingroup$ @RohitShukla : but when the matrix is not symmetric, it is not true anymore, take $M = \left(\begin{array}{ll}1&2000000\\0&0\end{array}\right)$ whose eigenvalues are $1$ and $0$, hence its spectral radius is $1$ ... $\endgroup$
    – reuns
    Commented Mar 29, 2016 at 8:11

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