Prove $\mathrm{span}(T ) = \mathrm{span}(T \cup \{ 0 \} ) $ I want to prove that

$$\mathrm{span}(T) = \mathrm{span}(T \cup \{ 0 \} ) $$

Try:
Pick some $x \in \mathrm{span}(T)$, then can write $x = \sum_{i=1}^n \alpha_i t_i $ where $t_i \in T$. Notice, we can write
$ x = \sum_i \alpha_i t_i + \alpha_{n+1} 0 $, then $x \in \mathrm{span}(T \cup \{ 0 \} $)
Isnt just the other direction the same argument>?

Yes, the other direction is identical!
Although, one could use a well known formula regarding sums of subspaces, that $\text{sp}(U\cup  W)=\text{sp}(U)+\text{sp}(W)$ when $W=\left \{ 0 \right \}$.
