This is just a question that I've been thinking about, but not sure what to do: From Naylor and Sell, Let X and Y be normed linear spaces and let T be a bounded linear transformation of X into Y. We define the norm of T to be
$||T||$=inf {${M:||Tx||_Y \leq M||x||_X \mbox{for all x} \in X}$}. Let X=Y=$R^{n}$ with norm $||x||_1$. I guess then we have that (not entirely sure):
- $||T||$=inf {${M:||Tx||_1 \leq M||x||_1 \mbox{for all x} \in X}$}
Then, we are also given that $M_n$ is the space of real n x n matrices. For
- $A=(a_{ij}) \in M_n$, we define $n(A)$=$\sum_{i,j}^{} |a_{ij}| $.
My question is then when is 1 equal to 2 and how do they relate. I know that $||T||$ is the operator norm, or the matrix induced norm by the vector 1-norm. Hence it is equal to the maximum absolute column sum. I believe $n(A)$ is just the sum of the absolute value of all the entries. So wouldn't this just imply that $n(A)\geq ||T|| $, with equality only occurring when we have one column with nonzero entries and the rest of the columns all having 0 as entries. The problem I'm having is whether I'm interpreting $||T||$ and $n(A)$ correctly. I'm not too familiar with this notation:$\sum_{i,j}^{} |a_{ij}| $. Any help would be much appreciated.
Your reasoning is entirely correct. In general, to work with operator norms it s better to think of them as $$\|T\|=\sup\{\|Tx\|:\ \|x\|=1\}. $$