How to determine that a set of elements of a linear space of $2\times3$ matrices is a linearly independent set Let $E_1,E_2,E_3,E_4 \in \cal{M}_{2 \times 3}(\Bbb{K})$ be defined as follows.
$$E_1= \begin{bmatrix}                         
    1 & 0 & 0 \\
    0 & 0 & 0 \\
\end{bmatrix}, E_2= \begin{bmatrix}
    0 & 1 & 0 \\
    0 & 0 & 0 \\
\end{bmatrix},$$ $$E_3= \begin{bmatrix}
    0 & 0 & 1 \\
    0 & 0 & 0 \\
\end{bmatrix}, E_4= \begin{bmatrix}
    0 & 0 & 0 \\
    1 & 0 & 0 \\
\end{bmatrix}.$$
$a)$ prove that set $\{E_i\}_{i \in I}$ is a linearly independent set,
$b)$ show that any element of $M$ can be represented as a linear combination of $\{E_i\}_{i \in I}$.
where $I = \{1,2,3,4\}$.
 A: First of all, if statement $(b)$ is true, then it must be that $I = \{1,2,3,4,5,6\}$ where:
$$E_5= \begin{bmatrix}
    0 & 0 & 0 \\
    0 & 1 & 0 \\
\end{bmatrix}$$
$$E_6= \begin{bmatrix}
    0 & 0 & 0 \\
    0 & 0 & 1 \\
\end{bmatrix}$$

$(a):$ To determine if $\{E_i\}_{i \in I}$ is a linaerly independent set, we use the definition of linear independence. That is, $\{E_i\}_{i \in I}$ is a linearly independent set if and only if:
$$a_1E_1 + a_2E_2 + a_3E_3 + a_4E_4 +a_5E_5 +a_6E_6 = \begin{bmatrix}
    0 & 0 & 0 \\
    0 & 0 & 0 \\
\end{bmatrix}$$
implies that $$a_1=a_2=a_3=a_4=a_5=a_6=0$$.
Thus, suppose that
$$A= \begin{bmatrix}
    a_1 & a_2 & a_3 \\
    a_4 & a_5 & a_6 \\
\end{bmatrix}= a_1E_1 + a_2E_2 + a_3E_3 + a_4E_4 +a_5E_5 +a_6E_6 =  \begin{bmatrix}
    0 & 0 & 0 \\
    0 & 0 & 0 \\
\end{bmatrix}$$
It is then clear that $a_1=a_2=a_3=a_4=a_5=a_6=0$ and thus $\{E_i\}_{i \in I}$ is a linearly independent set.

(b) Let A be a matrix in $\cal{M}_{2 \times 3}(\Bbb{K})$. Then we have that:
$$A= \begin{bmatrix}
    a_1 & a_2 & a_3 \\
    a_4 & a_5 & a_6 \\
\end{bmatrix}$$
For some $a_i \in (\Bbb{K})$.
Now, we define a linear combination of $\{E_i\}_{i \in I}$ so that the coefficient of $E_i$ in the linear combination is the entry $a_i$ of the matrix $A$:
$$a_1E_1 + a_2E_2 + a_3E_3 + a_4E_4 +a_5E_5 +a_6E_6 = \begin{bmatrix}
    a_1 & a_2 & a_3 \\
    a_4 & a_5 & a_6 \\
\end{bmatrix} = A$$
Since $A$ was an arbitrary $2 \times 3$ matrix, $(b)$ holds.
A: Let $V$ be a vector space over a field $\Bbb{K}$ and $\{v_1,v_2,\dots,v_k\} \subseteq V$, where $k \in \Bbb{N}$. We say that the set $\{v_1,v_2,\dots,v_k\}$ is linearly independent, if the null vector $0_V$ can only be expressed as a linear combination of $v_1, v_2, \dots, v_k$ with all the coefficients null. So, we say that this set is linearly independent if
\begin{align*}
\alpha_1 v_1 + \alpha_2 v_2 + \dots + \alpha_k v_k = 0_V \implies \alpha_1 = \alpha_2 = \dots = \alpha_k = 0.
\end{align*}
So, to answer the first question you have to start by writing the matrix $\cal{O}_{2 \times 3}$ as a linear combinations of $E_1, E_2, E_3, E_4$.
\begin{align}
\mathcal{O}_{2 \times 3} = \alpha_1 E_1 + \alpha_2 E_2 + \alpha_3 E_3 + \alpha_4 E_4.
\end{align}
This will give you (by solving the associated linear system), that $\alpha_1 = \alpha_2 = \alpha_3 = \alpha_4 = 0$. Therefore $\{E_i\}_{i \in I}$ is a linearly independent set.
For the second question, I think that there is a mistake. In fact you can’t write every matrix of $\cal{M}_{2 \times 3}$ as a linear combination of the elements of $\{E_i\}_{i \in I}$. Note that $\dim{\cal{M}_{2 \times 3}(\Bbb{K})} = 6$. Therefore, $\langle E \rangle \ \neq \cal{M}_{2 \times 3}(\Bbb{K})$. For example, how would you write the matrix
\begin{align*}
A = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}
\end{align*}
as a linear combinations of elements in $\{E_i\}_{i \in I}$?.
