Relation between Frobenius norm and L2 norm? Is there any relation between the Frobenius norm of a matrix and L2 norm of the vectors contained in this matrix.
Simply put, is there any difference between minimizing the Frobenius norm of a matrix and minimizing the L2 norm of the individual vectors contained in this matrix ?
Please help me understand this.
 A: Let $A$ be an $m\times n$ matrix with rows $\boldsymbol{a}_1^T,\dots,\boldsymbol{a}_m^T$. That is
$$A=
\begin{bmatrix}
\rule[.5ex]{1.5em}{0.4pt} & \boldsymbol{a}_1^T &\rule[.5ex]{1.5em}{0.4pt} \\
\rule[.5ex]{1.5em}{0.4pt} &\boldsymbol{a}_2^T& \rule[.5ex]{1.5em}{0.4pt} \\
&\vdots& \\
\rule[.5ex]{1.5em}{0.4pt} & \boldsymbol{a}_m^T &\rule[.5ex]{1.5em}{0.4pt}
\end{bmatrix}$$
The squared Frobenius norm is 
$$\|A\|_F^2=Tr(AA^T)=Tr\left(
\begin{bmatrix}
\boldsymbol{a}_1^T\boldsymbol{a}_1 & \boldsymbol{a}_1^T\boldsymbol{a}_2& \dots & \boldsymbol{a}_1^T \boldsymbol{a}_m \\
\vdots & \vdots & \ddots & \vdots\\
\boldsymbol{a}_m^T\boldsymbol{a}_1 & \boldsymbol{a}_m^T\boldsymbol{a}_2& \dots & \boldsymbol{a}_m^T \boldsymbol{a}_m
\end{bmatrix}
\right)=\sum_{i=1}^m\boldsymbol{a}_i^T\boldsymbol{a}_i=\sum_{i=1}^m\|\boldsymbol{a}_i\|_2^2$$
A: Consider the following matrix:
$$A=\begin{bmatrix}1 &0\\0&1\end{bmatrix}$$
The Frobenius norm is:
$$||A||_F=\sqrt{1^2+0^2+0^2+1^2}=\sqrt 2$$
But, if you take the individual column vectors' L2 norms and sum them, you'll have:
$$n=\sqrt{1^2+0^2}+\sqrt{1^2+0^2}=2$$
But, if you minimize the squared-norm, then you've equivalence. It's explained in the @OriolB answer.
A: Given a matrix $A \in \mathbb{K}^{n \times m}$, the squared Forbenius norm is defined as:
$||A||_F^2 := \sum_i^n \sum_j^m(a_{ij})^2 = Tr(A^TA)$,
observe the similarity to the squared $l_2$ norm for a vector $a \in \mathbb{K}^n$:
$||a||_2^2 := \sum_i^n(a_i)^2 = a^Ta$,
so we can rewrite the Frobenius norm as sum of squared $l_2$ norms of the vectors:
$||A||_F^2 = \sum_j^M ||a_j||_2^2 = \sum_j^M a_j^Ta_j$,
where $a_j \in \mathbb{K}^n$ are the column vectors of $A$. Thus if you minimize the squared norm it is equivalent.
