Let M $\in \mathbb{R}^{n\times n}$ with its norm given by $\|{M}\|$. I am trying to find a suitable upper bound of this matrix.

Is this inequality correct?

$M \leq \|M\|\mathbb{I}$, where $\mathbb{I} \in \mathbb{R}^{n\times n}$ is a matrix of all 1s and the inequality is implied entry wise.


1 Answer 1


Yes, that inequality is correct.

Suppose at least one entry of $M$ is larger than $\|M\|$, i.e. $M_{i,j} > \|M\|$ for some $i,j$.

Let $e_1,e_2, \ldots, e_n$ be the Euclidean basis vectors, i.e. the $k$-th entry of $e_k$ is $1$ and all the other entries are $0$.

Then, $\|Me_j\| = \|M_{i,j}e_i\| = |M_{i,j}| \ge M_{i,j} > \|M\| = \|M\| \cdot \|e_j\|$. This is a contradiction, since the matrix norm satisfies $\|Mx\| \le \|M\| \cdot \|x\|$ for all vectors $x$.

Therefore, $M_{i,j} \le \|M\|$ for all $i,j$, as desired.


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