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: 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 $1$.
A: If by "singular values" you mean the diagonal entries in writing $A=k_1 \pmatrix{t&0 \cr 0&t^{-1}}k_2$, with $k_1, k_2$ in the orthogonal group $SO(2)$, then your speculation is correct. EDIT: if you truly mean "spectral norm", then this is not correct. But/and in many contexts "singular values" does not refer to eigenvalues. Dunno... Let me continue under the assumption that you do not actually mean eigenvalues or spectral norm, but as I wrote... (someone may be interested in that scenario, in any case).
EDIT: this asserts nothing about diagonalizability of $A$. The general version is "Cartan decomposition". Then, letting $\delta=\pmatrix{a&0\cr 0&d}$,
$$
A^\top A \;=\; (k_1 \delta k_2)^\top(k_1Ak_2)
\;=\; k_2^\top \delta^\top k_1^\top k_1 \delta k_2
\;=\; k_2^\top \delta^2 k_2
$$
There are certainly variations depending on whether one wants to consider complex matrices, unitary versus orthogonal, and so on.
From such an expression, visibly, the singular values (EDIT: if this is truly what is intended) of the square are the squares of the singular values. For $SL_2(\mathbb R)$, the two singular values must be $t$ and $t^{-1}$.  Whichever is larger, when squared, is the larger of the two squares.  Since the maximum of $t$ and $t^{-1}$ is always at least $1$, the claim follows.
(Thanks to people trying to improve my answer, etc.)
EDIT-EDIT: (as @Jeppe Stig Nielsen notes in a comment), indeed, the operator norm is the square root of the largest eigenvalue of $A^\top A$, etc. Which is not so much related to the eigenvalues of $A$ itself, but to its "singular values" in an Iwasawa decomposition (see above). Perhaps this observation will resolve some confusion. (Believe me, these seemingly-slightly-different versions of similar things confounded me for a long time... :)
