Operator Norm = 1 Let there be a linear map $T$ such that $T: \mathbb R^n\to\mathbb R^m$. The operator norm $\lVert \cdot\lVert_{op}$ of $T$ is then defined as the largest value of $c$ for which $\lVert T(\vec v)\lVert \leq c\cdot\lVert\vec v\lVert,\;\forall\vec v\in\mathbb R^n$. In this case I'm interested in the cases where $\lVert\cdot\lVert$ is the Euclidean Norm. Provided that $T$ can be given in the form of a matrix $\mathbf A$ of dimension $m\times n$, or in other words provided that $T(\vec v) = \mathbf A\cdot\vec v$, it can be proven that
$$\lVert \mathbf A\lVert_{op} = \sqrt{\lambda_{max}}$$
where $\lambda_{max}$ is the largest eigenvalue of $\mathbf A^T\mathbf A$.
In the case where $\lVert \mathbf A\lVert_{op}\leq1$, we have that $\lambda_{max}\leq1$. What properties does $\mathbf A$ need to have in order to respect this last equation? Or in other words, what kind of proprieties does $\mathbf A$ need to have in order to have the largest value of $\mathbf A^T\mathbf A$ be less than or equal to 1?
Much appreciated.
 A: Note first that your definition of the operator norm is not correct. It should say that $\|T\|_{op}$ is the least $c$ such that $\|T\vec v\|\leq c\,\|\vec v\|$ for all $\vec v$.
Note also that the operator norm depends on which norms you give to $\mathbb R^n$ and $\mathbb R^m$. The canonical choice is to take the Euclidean norm in both, but that choice is by no means the only one. 
As for those $A$ such that $\|A\|_{op}\leq1$, you cannot expect any algebraic property to characterize them. Indeed, given any matrix $B$, the matrix $A=B/\|B\|_{op}$ satisfies $\|A\|_{op}=1$. 
Of course there are some necessary conditions. If $\|A\|_{op}\leq1$, then for all $k,j$
$$|A_{kj}|=|\langle Ae_j,e_k\rangle|\leq\|A\|_{op}\,\|e_j\|\,\|e_k\|\leq1.$$ More than that, if $e$ is the vector all entries equal to 1, then
$$
1\geq\|A\|_{op}\geq\,\frac1n\,|\langle Ae,e\rangle|=\frac1n\,\left|\sum_{k=1}^m\sum_{j=1}^nA_{kj}\right|.
$$
Or, for each $j$,
$$
1\geq\|A\|_{op}^2\geq\|Ae_j\|^2=\sum_{k=1}^m|A_{kj}|^2
$$
