I have seen that, in order to prove
$$ ||A||_{2} \leq ||A||_{F}$$
this inequality holds, proofs will use the following definition for a Frobenius norm.
$$||A||_{F}^{2} = \sum_{j=1}^{n} ||Ae_{j}||_{2}^{2}$$
However, I do not quite see how this is true. Can someone show me how this equality holds?
Note that $Ae_j$ is just the $j$-th column of $A$. Hence $\left\|Ae_j\right\|_{2}^{2}$ is the sum of squares of entries of the $j$-th column of $A$. So summing this for all $j$ gives us the sum of squares of all entries of the matrix, which is just $\left\|A\right\|^{2}_{F}$.
If you want to be more formal, write the $i,j$ entry of $A$ as $a_{ij}$ for $i=1,\ldots,m$ and $j=1,\ldots,n$ (where we assume $A$ is $m\times n$), then note that $Ae_j = \begin{bmatrix}a_{1j}\\ \vdots \\ a_{mj}\end{bmatrix}$. So $\left\|Ae_j\right\|_{2}^{2}=\sum\limits_{i=1}^{m}|a_{ij}|^{2}$. Therefore, $$\begin{align*} \sum\limits_{j=1}^{n}\left\|Ae_j\right\|_{2}^{2} &= \sum\limits_{j=1}^{n}\sum\limits_{i=1}^{m}|a_{ij}|^{2} \\ &= \left\|A\right\|_{F}^{2}. \end{align*}$$