The operator norm of a shear matrix

Idea. Compute $A^TA$, find its largest eigenvalue, and take a square root; then simplify the nested radical.

Step 1: the matrix $A^TA$.
$$A^TA=\begin{pmatrix}1&0\\1&1\end{pmatrix}\begin{pmatrix}1&1\\0&1\end{pmatrix}=\begin{pmatrix}1&1\\1&2\end{pmatrix}.$$

Step 2: eigenvalues. Its trace is $3$ and its determinant is $1$, so the characteristic polynomial is $t^2-3t+1$, with roots $\mu=\dfrac{3\pm\sqrt5}2$. The largest is $\mu_1=\dfrac{3+\sqrt5}2$.

Step 3: simplify the square root. Since
$$\left(\frac{1+\sqrt5}2\right)^2=\frac{1+2\sqrt5+5}4=\frac{3+\sqrt5}2,$$
we get $\|A\|=\sqrt{\mu_1}=\dfrac{1+\sqrt5}2\approx1.618$, the golden ratio.

Check. The singular values are $\varphi=\frac{1+\sqrt5}2$ and $\varphi^{-1}=\frac{\sqrt5-1}2$, with product $1=|\det A|$, and $\sigma_1^2+\sigma_2^2=3=\|A\|_F^2$. A direct numerical computation of the largest value of $|Ax|$ on the unit circle also gives $1.618\ldots$

Remark. The shear has $\rho(A)=1$ (both eigenvalues $1$) but $\|A\|=\varphi>1$; and $\|A^k\|$ grows linearly in $k$ (the matrix $A^k=\begin{pmatrix}1&k\\0&1\end{pmatrix}$), although $\|A^k\|^{1/k}\to1=\rho(A)$.