Is taking the Euclidean norm of multiple Euclidean norms equivalent to taking the Frobenius norm?

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I'm just a programmer venturing into the world of norms (is that even a thing?) here, and am wondering if two formulas are equivalent. Please forgive my ignorance!

Suppose we have a $10\times3$ matrix $\boldsymbol{X}$ ($10$ rows, $3$ columns).

Formula 1: The Euclidean norm of the Euclidean norms of the rows of $\boldsymbol{X}$

$$e = \sqrt{\sum_{i=1}^{10}{\|\boldsymbol{X_i}\|^2}}$$

Procedural description: take the Euclidean norm of all the rows of $\boldsymbol{X}$ and construct a new $10\times1$ matrix (so actually a $10$-element vector) containing the norms and call it $\boldsymbol{e^\prime}$. Then, take the Euclidean norm of $\boldsymbol{e^\prime}$ and call the final result $e$.

Formula 2: The Frobenius norm of $\boldsymbol{X}$

$$f = \|\boldsymbol{X}\|_F$$

Procedural description: Simply take the Frobenius norm of $\boldsymbol{X}$ and call the outcome $f$.


According to a number of simulations I've done on my machine, $e$ is consistently equal to $f$. Am I just getting 'lucky' or are the two actually equivalent?

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Good observation skills!

Let's chase definitions and then be happy :)

Let me write $n = 10, m = 3$ and let your matrix be denoted $A$. Then we calculate!

$$ \begin{align} f &= \|A\|_F \;\; \text{ your defn} \\ & = \sqrt{\sum_{i=1}^n\sum_{j=1}^m |A_{ij}|^2}\;\; \text{ defn of Frobenius norm} \\ & = \sqrt{\sum_{i=1}^n\sqrt{\sum_{j=1}^m |A_{ij}|^2}^2} \;\; \text{ maths: $\sqrt{x}^2=x$ if $x \geq 0$} \\ & = \sqrt{\sum_{i=1}^n \|A_i\|^2} \;\;\text{ defn of Euclidean norm of row $i$} \\ & = e \;\; \text{ your defn } \end{align} $$

Yay :)

Hope that helps!

(See http://en.wikipedia.org/wiki/Matrix_norm)