Show that the Frobenius norm $\| \cdot \|_{F} : \mathbb{K}^{n \times n} \to \mathbb{R}_{+}$ is a matrix norm on $\mathbb{K}^{n \times n}$ and is submultiplicative and compatible to the vector norm $\|\cdot\|_{2}: \mathbb{K}^{n} \to \mathbb{R}_{+}$.
Hint: Use the Cauchy-Schwarz inequality $$\left(\sum_{i=1}^{n} u_{i} v_{I} \right)^{2} \leq \left(\sum_{j=1}^{n} \left|u_{j} \right|^{2}\right) \cdot \left(\sum_{k=1}^{n} \left|v_{k}\right|^{2}\right)$$
What I've researched is the following. The Frobenius norm is:
$$\|A\|_{F}:=\sqrt{\sum_{i=1}^{m} \sum_{j=1}^{n}\left|a_{i j}\right|^{2}}$$
The Euclidean norm is:
$$\|x\|_{2}=\sqrt{\sum_{i=1}^{n}\left|x_{i}\right|^{2}}$$
Both has to be set up like that:
$$ \|A \cdot x\|_{V} \leq\|A\| \cdot\|x\|_{V} $$
Like the hint. So, first off, I don't get how to get to the hint starting with these norms. Need some help to understand the way of doing that.
Denote the columns of $A$ as $a_i$, $1\leq i\leq n$. Then we can rewrite the Frobenius norm as $$\|A\|_F^2 = \sum_{i=1}^n\|a_i\|_2^2. $$
To show these results, it is easiest to use the fact that the Frobenius norm is invariant under multiplication with orthonormal columns from the left and invariant under multiplication with orthonormal rows from the right to get a simple inequality between the Frobenius norm and matrix $2$-norm. Using the SVD of $A$, we have $\|A\|_F^2 = \|U\Sigma V^*\|_F^2=\|\Sigma\|_F^2 = \sum_{i=1}^{n}\sigma_i^2$. If we recall that $\|A\|_2^2 = \sigma_1^2$, then we have $\|A\|_2\leq\|A\|_F$.
We now show that the Frobenius norm is submultiplicative.
Given $A$ and $B$ and using column notation as before, we have $$ \|AB\|_F^2=\sum_{i=1}^n\|Ab_i\|_2^2\leq\|A\|_2^2\sum_{i=1}^n\|b_i\|_2^2=\|A\|_2^2\|B\|_F^2\leq\|A\|_F^2\|B\|_F^2, $$ and we are done.
To show compatibility with the vector $2$-norm, we use submultiplicativity of the matrix $2$-norm: $$ \|Ax\|_2\leq\|A\|_2\|x\|_2\leq\|A\|_F\|x\|_2 $$ and we are done.