$\DeclareMathOperator\SL{SL}\DeclareMathOperator\trace{trace}$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 to Is it true that $\lVert A\rVert \leq \lVert A^2\rVert$ for $A\in \operatorname{SL}(2, \mathbb{R})$? 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.

`A^{^*}`

for the adjoint; it should be $A^*$`A^*`

. I edited accordingly. $\endgroup$