Link between the norm $1$ of a matrix and its biggest eigenvalue I am working on a set of matrices for a project, studying their highest eigenvalue, let's call it $\lambda_{1}$.
I was curious and plotted the norm 1 of the matrix, ie $ \frac{1}{n^{2}}\sum_{i,j} |a_{i,j}|$.
I observed a similar comportement for these quantities, is there any explanation ?
Is this due to the structure of my matrices ? Is there a general results ?
Thanks in advance
More information about my matrix: It's a correlation matrix, symetric and positive.
The first eigenvalues is approximately $10$ times bigger than the others.
I also forgot an important thing: all variables are normalized. (and also the norm 1)

$\lambda1$ in red, the norm 1 in green.
 A: Edit: It is unsurprising that the spectral radius $\rho(A)$ and the $\ell_1$-norm of $A$ have roughly the same trends. Since $A$ is positive definite, we have $\rho(A)=\|A\|_2$. As all norms on a finite dimensional vector space are equivalent, we would expect $\rho(A)$ to be large when the $\ell_1$-norm of $A$ is large, and vice versa.
What is surprising, though, is that $\rho(A)$ and the normalized $\ell_1$-norm of $A$ have roughly the same magnitudes. By spectral decomposition, $A=\sum_j \lambda_j \mathbf{v}_j\mathbf{v}_j^T$, where each $\mathbf{v}_j$ is a unit eigenvector of $A$ corresponding to the eigenvalue $\lambda_j$, with $\lambda_1\ge\lambda_2\ge\cdots\ge\lambda_n$. Now:


*

*If $\lambda_1=\cdots=\lambda_n$, i.e. $A=\lambda_1I$, we have $\frac1{n^2}\sum_{i,j}|a_{ij}|=\frac{\lambda_1}n$.

*If $\lambda_2,\ldots,\lambda_n\ll\lambda_1$, then $A\approx\lambda_1 \mathbf{v}_1\mathbf{v}_1^T$ and
$$
\frac1{n^2}\sum_{i,j}|a_{ij}|=\frac1{n^2}\mathbf{1}^T|A|\mathbf{1}
\approx\frac{\lambda_1}{n^2}\langle \mathbf{1},|\mathbf{v}|\rangle^2.
$$
Yet, by Cauchy-Schwarz inequaltiy, $\langle \mathbf{1},|\mathbf{v}_1|\rangle^2\le\|\mathbf{1}\|^2\ \|(|\mathbf{v}_1|)\|^2=n$. Therefore we would expect the order of magnitude of $\frac1{n^2}\sum_{i,j}|a_{ij}|$ to be not far greater than $\frac{\lambda_1}n$.


Now you say that  $\lambda_1$ is roughly ten times the other eigenvalues. While $\lambda_2,\ldots,\lambda_n$ are an order of magnitude smaller than $\lambda_1$, since there are $O(n)$ of them, they collectively may be non-negligible. However, based on the above two example cases, I would expect the order of magnitude of $\frac1{n^2}\sum_{i,j}|a_{ij}|$ to be $\frac{\lambda_1}n$ rather than $\lambda_1$. So, your findings are quite unexpected. Perhaps you may repeat the same numerical experiment for larger-sized $A$, and see if $\rho(A)$ and the normalized $\ell_1$ norm still have roughly the same magnitude. If they do, there is a good chance that there's really something interesting in it; if they don't, then your current findings are perhaps mere coincidence.
A: It is surely cause of the structure of your matrix. Only for $1\times 1$ matrices is it true in general.
Take any upper triangular matrix, you can chose the values on the upper triangular matrix as large as you like still all eigenvalues will be $0$.
If you like more complicated counterexamples, take $u,v\in \mathbb{R}^n$ and study the matrix 
$B=u\cdot v^T$ with $u\neq 0$ and $ v\bot u$. Here all eigenvalues are $0$, and the 
$\infty$ norm of $B$ is the $\infty$ norm of $u$ times the 1 norm of $v$.
For symmetric positiv definit matrices the largest eigenvalue can't be bigger than the $1$ norm, as the trace is invariant under basistransformation and as diagonal form there are only positive values in the diagonal (as your matrix is positiv definit). 
