Eigenvalues are bounded by the operator norm

Idea. For a real eigenvalue the claim is immediate from $|Ax|\le\|A\||x|$. For a non-real one, we cannot apply the definition of $\|A\|$ (a supremum over *real* vectors) directly, so we split the complex eigenvector into its real and imaginary parts.

Step 1: real eigenvalues. If $Ax=\lambda x$ with $x\in\mathbb{R}^n\setminus\{0\}$, then $|\lambda||x|=|Ax|\le\|A\||x|$, so $|\lambda|\le\|A\|$.

Step 2: general case. Let $v=x+iy\ne0$ with $x,y\in\mathbb{R}^n$ and $Av=\lambda v$. The Hermitian norm is $|v|^2=|x|^2+|y|^2$. Since $A$ is real, $Av=Ax+iAy$ with $Ax,Ay\in\mathbb{R}^n$, so
$$|\lambda|^2(|x|^2+|y|^2)=|Av|^2=|Ax|^2+|Ay|^2\le\|A\|^2|x|^2+\|A\|^2|y|^2=\|A\|^2(|x|^2+|y|^2).$$
Dividing by $|x|^2+|y|^2>0$ gives $|\lambda|\le\|A\|$.

Example. For the rotation $R_\theta$, the eigenvalues are $e^{\pm i\theta}$ with $|\lambda|=1=\|R_\theta\|$: the bound is sharp. For $A=\begin{pmatrix}0&1\\0&0\end{pmatrix}$, $\rho=0<1=\|A\|$.

Remark. The same proof works for every induced norm provided it is extended to complex vectors; the Euclidean case is the easiest because the Hermitian norm splits as above. This inequality is the basic step towards the Gelfand bound $\rho(A)\le\|A^k\|^{1/k}$.