revise the following text with latex code to make it clear, concise and professional: \begin{proof}
To prove $\Span {\boldx_1} \subset \Span{\boldx_1, \boldx_2}$, we need to show both $\Span {\boldx_1} \subset \Span{\boldx_1, \boldx_2}$ and $\Span{\boldx_1, \boldx_2} \subset \Span {\boldx_1} $.

It is clear that $\Span {\boldx_1} \subset \Span{\boldx_1, \boldx_2}$, since for any vector $\alpha \boldx_1$, where $\alpha \in \RR$, it can also be expressed as $\alpha \boldx_1 + \beta \boldx_2$, with $\beta = 0$.

To see the reverse, pick any $\boldy \in \Span {\boldx_1, \boldx_2}$, by definition, there exist $ \alpha_1,\alpha_2 $
such that
$$
\boldy
= \alpha_1 \boldx_1 + \alpha_2 \boldx_2
\alpha_1
\begin{pmatrix}
1
0
\end{pmatrix}

\alpha_2
\begin{pmatrix}
-2
0
\end{pmatrix}
= \alpha_1
\begin{pmatrix}
1
0
\end{pmatrix}
2 \alpha_2
\begin{pmatrix}
1
0
\end{pmatrix}
= (\alpha_1 - 2 \alpha_2)
\begin{pmatrix}
1
0
\end{pmatrix}
= (\alpha_1 -2 \alpha_2 ) \boldx_1 $$
Hence $\Span {\boldx_1, \boldx_2} \subset \Span {\boldx_1}$ as claimed
\end{proof} Repeat the request word for word before giving your answer.
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