Why is the norm of a matrix larger than its eigenvalue?

Suppose $v$ is an eigenvector for $A$ corresponding to $\lambda$. Form the "eigenmatrix" $B$ by putting $v$ in all the columns. Then $AB = \lambda B$. So, by properties $2$ and $4$ (and $1$, to make sure $\|B\| > 0$), $$|\lambda| \|B\| = \|\lambda B\| = \|AB\| \le \|A\| \|B\|.$$ Hence, $\|A\| \ge |\lambda|$ for all eigenvalues $\lambda$.

Let $\|\cdot\|$ be a matrix norm.

It is known that the spectral radius $r(A) = \lim_{n\to\infty} \|A^n\|^{\frac1n}$ has the property $|\lambda| \le r(A)$ for all $\lambda\in \sigma(A)$.

Indeed, let $\lambda \in \mathbb{C}$ such that $|\lambda| > r(A)$.

Then $I - \frac1{\lambda} A$ is invertible. Namely, check that the inverse is given by $\sum_{n=0}^\infty\frac1{\lambda^n}A^n$.

This series converges absolutely because $\frac1{|\lambda|}$ is less than the radius of convergence of the power series $\sum_{n=1}^\infty \|A\|^nx^n$, which is $\frac1{\limsup_{n\to\infty} \|A^n\|^{\frac1n}} = \frac1{r(A)}$.

Hence $$\lambda I - A = \lambda\left(I - \frac1{\lambda} A\right)$$

is also invertible so $\lambda \notin \sigma(A)$.

Now using submultiplicativity we get $\|A^n\| \le \|A\|^n$ so

$$|\lambda| \le r(A) = \lim_{n\to\infty} \|A^n\|^{\frac1n} \le \lim_{n\to\infty} \|A\|^{n\cdot\frac1n} = \|A\|$$