$\DeclareMathOperator\SL{SL}\DeclareMathOperator\trace{trace}$Let $A \in \SL(2,\mathbb{R})$ and $\trace(A)>2$. Is it true that $$\lVert A\rVert \leq \lVert A^2\rVert,$$
where $\lVert \rVert$ is the operator norm that is the first singular value?
$$\lVert A \rVert =\sqrt{\lambda\_{\text{max}}(A^\*A)}=\sigma\_{\text{max}}(A).$$


Let me mention that if the condition $\trace(A)>2$ is removed, then the above statement is not true; see Jeppe Stig Nielsen's [answer](https://math.stackexchange.com/a/4282670) to [Is it true that $\lVert A\rVert \leq \lVert A^2\rVert$ for $A\in \operatorname{SL}(2, \mathbb{R})$?](https://math.stackexchange.com/questions/4282612/is-it-true-that-a-leq-a2-for-a-in-sl2-mathbbr) on MSE.


My attempt:
I think it is true: The operator norm satisfied
$$\lVert A\rVert=\sup\left\{\lVert Ax\rVert \,\middle\vert\, \text{$x\in\mathbb{R}^2$ and $\lVert x\rVert=1$}\right\}$$
where the symbols $\lVert \rVert$ inside the brackets on the right-hand side denote the standard (Euclidean) length of a vector in $\mathbb{R}^2$. So $\lVert A\rVert$ is the maximal length of the image of a unit vector. On the other hand, $\trace(A)>2$, so one of the eigenvalues is greater than the other one.



Is it true that $\lVert A\rVert \leq \lVert A^2\rVert$ for $A\in \operatorname{SL}(2, \mathbb{R})$ when $\operatorname{trace}(A)>2$?
We can do this by a calculation. The assumptions on the determinant and trace are equivalent to having eigenvalues $\lambda,1/\lambda$, with $\lambda>1$. We can rotate the first eigenvector to the $e\_1$ position, and then
$$
A=\begin{pmatrix} \lambda & b \\ 0 & 1/\lambda \end{pmatrix} ,
$$
so
$$
A^\*A=\begin{pmatrix} \lambda^2 & \lambda b \\ \lambda b & b^2 +1/\lambda^2
\end{pmatrix}
$$
In general, the eigenvalues of a $B\in\textrm{SL}(2,\mathbb R)$ are $T/2 \pm \sqrt{T^2/4-1}$, with $T=\textrm{tr}\; B$.


In the case of $A^\*A$, we are interested in the large eigenvalue (obtained for $+$). Clearly, this is increasing in $T$. So now the question is if $(A^2)^\*A^2$ has a larger trace $T\_2$, and the same calculation now gives
$$
T\_2 = \lambda^4 + \frac{1}{\lambda^4} + b^2 \left( \lambda+\frac{1}{\lambda}\right)^2 .
$$
This indeed satisfies $T\_2\ge \lambda^2+1/\lambda^2 + b$. In fact, we have $\lambda^4+1/\lambda^4\ge \lambda^2+1/\lambda^2$ for $\lambda\ge 1$, or, equivalently, $f(x)\equiv x^4-x^3-x+1\ge 0$ for $x\ge 1$, since $f'(x)\ge 0$ in this range and $f(1)=0$.


