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Why is the euclidean norm of a matrix (linear operator) is greater than the norm defined by maximim of the absolute value of entries of the matrix?

2 Answers 2

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If the largest entry of $A$ in absolute value is $a_{ij}$, then letting $E_j$ be the $j$th standard basis vector of the space $\mathbf{R}^n$ of column matrices, we have $|A| = |A||E_j| \geq |AE_j| \geq |a_{ij}|$.

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Edit:

If $\|A\|_2 = \sup_{\|x\|_2=1}{(\|Ax\|_2)}$ is the Euclidean norm (operator norm) then the following proof holds.

Let $A$ be a $m$x$n$ matrix. We know $\sup_{\|x\|_2=1}{(\|Ax\|_2)} \geq \|Ae_i\|_2$ for any $1 \leq i \leq n$, where $e_k$ is a column vector of size $n$x$1$ taking value $1$ at $k^{\text{th}}$ index and $0$ elsewhere.

$\|Ae_i\|_2$ is the Euclidean norm of $i^{\text{th}}$ column of matrix $A$, thus we have $\|Ae_i\|_2 \geq |a_{*i}|$.

For previous answer $ \|A\| $ is the Frobenius norm.

Hint: $ \|A\| = \left(\sum_{i=1}^m \sum_{j=1}^n |a_{ij}|^2\right)^{1/2} \geq |a_{pq}|$ for any $1 \leq p \leq m, 1 \leq q \leq n.$

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    That's the frobenius norm... It's the euclidean norm of the vectorization of the matrix. That's not, how I read that question. – 2017-02-17
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    Frobenius norm and Euclidean norm are interchangeable names for matrices. Refer http://mathworld.wolfram.com/FrobeniusNorm.html or any other source. – 2017-02-17
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    I always used "Euclidean Norm" for $\sup_{\|x\|_2=1}{(\|Ax\|_2)}$, the induced 2-Norm of the linear operator. – 2017-02-17
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    @Laray Thanks for the information. Updated the answer likewise. – 2017-02-17