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Let

$$A=\begin{bmatrix} {a_1}_1 & {a_1}_2 & \cdots & {a_1}_n\\ {a_2}_1 & {a_2}_2 & \cdots & {a_2}_n \\ \vdots & \vdots & \ddots & \vdots\\ {a_m}_1 & {a_m}_2 & \cdots & {a_m}_n \\ \end{bmatrix} = \begin{bmatrix} \vec a_1 & \vec a_2 & \cdots & \vec a_n\end{bmatrix}$$

The Frobenius norm of matrix $A$ is

$$||A||_F = \sqrt{\mathrm{tr} (A^\top A)} = \sqrt{\sum_{i=1}^m \sum_{j=1}^n |{a_i}_j|^2} \tag{1}$$

If I take $||A||_F^2$, can I write it as

$$||A||_F^2=\sum_{j=1}^n ||\vec a_j||^2 \tag{2}$$

Is equation $(2)$ mathematically right?

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    $\begingroup$ It's correct, although $a_i$ should be $\vec a_i$. $\endgroup$ – littleO Sep 14 '17 at 9:07
  • $\begingroup$ What have you tried? What is the result if you just plug in the definition of $\|a_i\|$? $\endgroup$ – P. Siehr Sep 14 '17 at 9:07
  • $\begingroup$ Sorry for my mistake, I just edited the eq.(2), $a_i$ is a vector; $\vec a_i$. $\endgroup$ – Muhammad Hammad Saghir Sep 14 '17 at 9:24
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Short answer is yes.

To see it, just look at what happens if you expand the definition of the vector norm in your equation (2) and take its square.

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    $\begingroup$ If you are going to answer a question, answer it; don't ask another question. $\endgroup$ – uniquesolution Sep 14 '17 at 9:06
  • $\begingroup$ Just putting OP on the rails to see why his answer is valid. $\endgroup$ – user70925 Sep 14 '17 at 9:08
  • $\begingroup$ I tried it with $2 \times 2$ matrix, it seems it is true but I just wanted to be sure and asked for your opinions for confidence. $\endgroup$ – Muhammad Hammad Saghir Sep 14 '17 at 9:27
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Just to make sure my notations are clear, let $$A=\begin{bmatrix} {a_1}_1 & {a_1}_2 & \cdots & {a_1}_n\\ {a_2}_1 & {a_2}_2 & \cdots & {a_2}_n \\ \vdots & \vdots & \ddots & \vdots\\ {a_m}_1 & {a_m}_2 & \cdots & {a_m}_n \\ \end{bmatrix} = \begin{bmatrix} | & | & & | \\ a_1 & a_2 & \cdots & a_n \\ | & | & & | \\ \end{bmatrix}$$ where $a_j \in \mathbb{R}^m$ for all $j = 1, \ldots, n$.

As was answered in this answer, you can check that you equation (2) is valid by computing: $$||A||_F^2 \overset{eq. (1)}{=} \sum_{i=1}^m \sum_{j=1}^n a_{ij}^2 = \sum_{j=1}^n \sum_{i=1}^m a_{ij}^2 = \sum_{j=1}^n ||a_j||_2^2$$

Here is another, an more visual, answer to this question. One common definition of the Frobenius norm is the following (form real matrices): $||A||_F := \sqrt{\mathrm{tr} (A^\top A)}$. So, to compute the squared Frobenius norm of $A$, one can sum the diagonal terms of $A^\top A$, ie, $$||A||_F^2 = \mathrm{tr} (A^\top A) = \sum_{j=1}^n (A^\top A)_{jj}$$

Let us then visualize what we have the diagonal of this matrix product:

\begin{align*} ~ & \, \begin{bmatrix} | & | & & | \\ a_1 & a_2 & \cdots & a_n \\ | & | & & | \\ \end{bmatrix} \\ % A^\top A = \begin{bmatrix} - & a_1^\top & - \\ & \vdots & \\ - & a_n^\top & - \\ \end{bmatrix} & \begin{bmatrix} ||a_1||_2^2 & * & * \\ * & \ddots & * \\ * & * & ||a_n||_2^2 \end{bmatrix} \end{align*}

So, when you compute the trace, you recover the sum of the squared norms.

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