If the 2-norm of a matrix is small, the trace of the matrix is also small Is it true that If the 2-norm of a symmetric real matrix is small, then the trace of the matrix is also small? I played around with some matrices in MATLAB and discovered this phenomenon. Does there exist any theorem that relates the 2-norm of a symmetric real matrix to the magnitude of its trace? 
Thanks.
 A: If the matrix $A$ is real symmetric it is diagonalizable. If $v$ is a unit eigenvector and $\lambda$ the corresponding eigenvalue, then 
$$|\lambda|=|\lambda v|_2=\|Av\|_2\le \|A\|_2.$$ 
That is, if the $2$-norm is small  then any of its eigenvalues is small. But the trace of the matrix is the sum of the eigenvalues, which is small, because any of the eigenvalues is small.
That is, we have the upper bound 
$$|\textrm{trace}(A)|=|\sum_{i=1}^n\lambda_i|\le \sum_{i=1}^n|\lambda_i|\le n\|A\|_2.$$
A: In addition to the inequality mfl gave, there is a rather intuitive way of answering this question.


*

*The trace is the sum of all eigenvalues (with multiplicities): $\text{tr}(A) = \sum_i \lambda_i$.

*The 2-norm of a symmetric matrix is the maximum eigenvalue: : $||A||_2 = \max_i |\lambda_i|$.


If you have negative and positive eigenvalues, the trace may arbitrarily small while the 2-norm is bounded (because the eigenvalues can cancel out, while still being of large magnitude).
On the other hand if the eigenvalues are roughly all the same, the absolute value of the trace may be $n$ times larger than the 2-norm. And this is just what mfl's inequality is ($|\text{tr}(A)|\leq n ||A||_2$).
The argument also shows that the inequality is sharp and really all cases can happen. It also tells how the eigenvalues have to be distributed for a certain case.
