Minimize $x^TAx$ with respect to $T$, where $A=T\odot T^{-1}$

Question

\begin{array}{ll} \text{minimize} & x^TAx\\\quad T\in\mathbb{R}^{n\times n}\\ \text{subject to} & A=T\odot T^{-1}\\&T>0\end{array}

where the symbol $\odot$ denotes the elementwise/Hadamard product, and $x_i>0.$

---

My attempt:

Here $T\odot T^{-1}$ is called Relative gain array. I found that $\lambda_{min}(A)\geq 1$ (on page 483, problem 7.5.P14), so $x^TAx\geq \lambda_{min}(A)x^Tx\geq x^Tx$. Then we need to construct $A$, such that $A^{\frac{1}{2}}x=x$. If we take $A=T=I$, then we achieve $x^TAx=x^Tx$.

Counter-example to my attempt:

x=[5 2]';
for i=1:500
   P=rand(2,2);
   T=P'P;
   d(i)=x'*(T.*inv(T))*x-x'*x;
end
plot(d)

$d$ here is strictly larger than zero.

---

Edit: there was an error in code. Now it is fixed. Looks like the $T=I$ is optimal.

Answer 1

Maybe will be useful for someone. If $A$ has all eigenvalues larger than $1$, then $x^TAx\geq x^Tx$, thus we need to find positive definite $T$, such that its relative gain array $T\odot T^{-1}=I.$ One property of relative gain array is that for any upper triangular $T$, its relative gain array is equal to identity matrix. Thus we can just choose diagonal $T$ with all diagonal elements strictly larger than zero.