Norm of matrix and its maximum eigenvalue I've seen in some inequalities in the theory of ODEs that $\lVert Q \lVert \le \lambda_{max}(Q)$. What theorem from Linear Algebra is relevant here?
 A: I wonder whether the inequality isn't backward. 
$\|A\|_2$ is usually defined as the supremum, over all unit vectors $x$, of $\|Ax\|_2$. 
Let $\lambda$ be the largest eigenvalue of $A$ (in modulus), and let $v$ be a corresponding unit eigenvector. Then $$\|A\|_2\ge\|Av\|_2=\|\lambda v\|_2=|\lambda|\|v\|_2=|\lambda|$$
A: The equation is backwards, but there is a more general result that is applicable. This result can be found in textbooks such as Theorem 5.6.9 in   Matrix Analysis (2012) by Horn and Johnsson:

Let $\|Q\|$ denote a matrix norm of the matrix $Q$. Let  $\lambda$ be
an eigenvalue of $Q$. Let $\rho(Q)$ be the spectral norm of $Q$, which
is the absolute value of the largest eigenvalue:  $\rho(Q) \triangleq |\lambda_{max}(Q)|$.
$$|\lambda| \leq \rho(Q) \leq \|Q\| $$
and if $Q$ is invertible, then  $$\rho(Q) \geq |\lambda| \geq 1/ \|Q\| $$

Remark: If $\lambda_{max}(Q)$ is a nonnegative real value, then $\lambda_{max}(Q)=\rho(Q)$, and $ \lambda_{max}(Q) \leq \|Q\|$. Example of such occasions is if $Q$ is positive semidefinite or if $Q$ is a irreducible nonnegative (see Perron-Frobenius theorem).
A: You can also use the Jordan Normal Form of the matrix to show that all of the eigenvalues are less than the norm of the matrix.       
