the matrix inequlity Is it true that $$\|A\| \leq \|A^2\|$$
for $A \in SL(2,\mathbb{R})$, where $\| \|$ is the operator norm that is the first singular value?
$$\left \| A \right \| =\sqrt{\lambda_{\text{max}}(A^{^*}A)}=\sigma_{\text{max}}(A).$$
Definitively, it is not true for $GL(2, \mathbb{R})$ as one can consider $A=diag(1/3, 1/3).$

A counterexample could be $$A=\pmatrix{3&5\\-2&-3}.$$

Some thought about how you can see that this $A$ works.
First of all, its determinant is $-9-(-10)=1$ as required. This also means that the product of its two eigenvalues is $1$.
The trace of $A$ is $3-3=0$. So the sum of the eigenvalues is $0$. Given the product and the sum of the eigenvalues, it is clear that the eigenvalues are $+i$ and $-i$ where $i$ is the imaginary unit.
Since the eigenvalues are distinct, this $A$ is diagonalizable by some complex matrix in $GL(2,\mathbb{C})$.
Now think about $A^2$. It is diagonalizable as well (because $A$ is, and by the same complex matrix). The eigenvalues of $A^2$ are $-1$ and $-1$, the squares of the eigenvalues of $A$. But a diagonalizable matrix all of whose eigenvalues are equal, is really a diagonal matrix, so we have $$A^2=\pmatrix{-1&0\\0&-1}.$$
The operator norm you are asking about satisfies $$\|A\|=\sup\left\{\|Ax\|\,\middle|\, x\in\mathbb{R}^2 \text{ and } \|x\|=1\right\}$$ where the symbols $\|\cdot\|$ inside the set on the right-hand side denotes the standard (Euclidean) length of a vector in $\mathbb{R}^2$. So $\|A\|$ is the maximal length of the image of a unit vector.
It is clear that $\|A^2\|=1$ since $A^2$ maps all unit vectors to unit vectors. It is also clear that $\|A\|$ is strictly larger. For simplicity, take $x=\pmatrix{0\\1}$, then $Ax=\pmatrix{5\\-3}$ whose length is $\sqrt{5^2+3^2}=\sqrt{34}$, so it follows that the operator norm of $A$ is at least this number, $\|A\|\ge\sqrt{34}$.

These arguments show that any matrix $$A=\pmatrix{t&s\\-\frac{1+t^2}{s}&-t}$$ where $s\ne 0$ and $t$ are real numbers has determinant $1$ and trace $0$ and therefore will square to $\pmatrix{-1&0\\0&-1}$. By taking both $s$ and $t$ huge, you can make the length of the second column of $A$ as huge as you want. So there are matrices with arbitrarily large operator norms in $SL(2,\mathbb{R})$ whose squares have operator norm $1$.
