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

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.

• Your accepted answer is incorrect, by the way. Cameron pointed out the flaw. – Josh Keneda Mar 29 '16 at 4:23
• @JoshKeneda It isn't. The question is not about the entries of $A$, but about the entries of a matrix $H$ that is constructed from $A$. The accepted answer is perfectly fine. – user1551 Mar 31 '16 at 10:35
• Ah. Good catch. I do wish that the accepted answer mentioned how crucial the normality of $H$ is in the argument, though. The title of the question seems misleading, since the key property was the norm bound on $H$. – Josh Keneda Apr 1 '16 at 0:06

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

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

• I am sorry for the redundant question but what is $e^{T}_{i}$ and $e_{j}$? – Rohit Shukla Mar 29 '16 at 2:59
• @RS $e_i$ is the (column) vector given by $$e_i = (\overbrace{0,\dots,0}^{i-1},1,0,\dots,0)$$ – Omnomnomnom Mar 29 '16 at 3:01
• Maybe I am mistaken, but you used $\|B\|\le 1$, not $\rho(B)\le 1$. If $\|B\|\le\rho(B)$, this would hold, but instead the opposite inequality holds. – Cameron Williams Mar 29 '16 at 4:04
• @Omnomnomnom: your proof is wrong. Try to apply it to Fnacool's matrix with $i=1,j=2$.3 If you use that $B$ is selfadjoint, you should say so. – Martin Argerami Mar 29 '16 at 23:19
• @MartinArgerami I see no flaws in this answer. The OP is about their $H$, not $A$. By construction, unless $A$ is non-real, $H$ is a real symmetric matrix with spectral norm (= spectral radius as $H$ is real symmetric) $\le1$. – user1551 Mar 31 '16 at 10:28

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 $\frac 12$.

• This answer is wrong. The OP is not about $A$, but about $H=I-\frac{2A^TA}{\operatorname{tr}(A^TA)}$. – user1551 Mar 31 '16 at 10:29