The operator norm via the largest eigenvalue of Aáµ€A

Idea. The quantity $|Ax|^2$ is a quadratic form in $x$ with matrix $S=A^TA$; the maximum of a quadratic form on the unit sphere is the largest eigenvalue of its (symmetric) matrix.

Step 1: $S=A^TA$ is symmetric PSD. $S^T=A^T(A^T)^T=S$, and $x^TSx=|Ax|^2\ge0$. So the spectral theorem gives an orthonormal eigenbasis $u_1,\dots,u_n$ with eigenvalues $\mu_1\ge\dots\ge\mu_n\ge0$ (non-negative since $\mu_i=u_i^TSu_i=|Au_i|^2$).

Step 2: maximum of the quadratic form. Write $x=\sum c_iu_i$ with $\sum c_i^2=|x|^2=1$. Then
$$|Ax|^2=x^TSx=\sum_i\mu_ic_i^2\le\mu_1\sum_ic_i^2=\mu_1,$$
with equality for $x=u_1$. Hence $\|A\|^2=\sup_{|x|=1}|Ax|^2=\mu_1$, so $\|A\|=\sigma_1=\sqrt{\mu_1}$.

Step 3: Frobenius norm. $\|A\|_F^2=\operatorname{tr}(A^TA)=\operatorname{tr}(S)$ and the trace of a symmetric matrix is the sum of its eigenvalues. So $\|A\|_F^2=\sum_i\mu_i=\sum_i\sigma_i^2$.

Remark. Consequently $\|A\|=\sigma_1\le\big(\sum\sigma_i^2\big)^{1/2}=\|A\|_F\le\sqrt n\,\sigma_1=\sqrt n\|A\|$, which reproves the comparison of the two norms, with equality cases visible: $\|A\|=\|A\|_F$ iff only one singular value is non-zero (rank $\le1$), and $\|A\|_F=\sqrt n\|A\|$ iff all singular values are equal (multiples of orthogonal matrices).