# $V$ is finite dimensional vector space. $G$ is a spanning set of $V$, $I \subseteq G$ is linearly independent

I'm totally new to the subjects like spanning set, bases and dimension. Here is a theorem that I could not understand.

Let $$V$$ be a finite dimensional vector space. If $$G$$ is a finite spanning set of $$V$$ and if $$I \subseteq G$$ is linearly independent, then there is a basis $$B$$ of $$V$$ such that $$I \subseteq B \subseteq G$$.

If $$\text{Span}(I)= V$$ then $$B = I$$ and we have nothing left to prove.

I know that if $$\text{Span}(I)= V$$ and since $$\text{Span}(B)= V$$ follows $$\text{Span}(I)= \text{Span}(B)$$ but how do we know $$I=B$$ and why we have nothing left to prove?

Suppose $$\text{Span}(I) \ne V$$ then $$I \ne G$$ and $$G \setminus \text{Span}(I) \ne \emptyset$$ (if $$G \subseteq \text{Span}(I)$$ we would have $$V=\text{Span}(G) \subseteq \text{Span}(I)$$).Thus $$\exists g_1 \in G$$ such that $$g_1 \notin \text{Span}(I)$$. Then $$I'= I \cup \left\{g_1 \right\} \subseteq G$$ is linearly independent.

Why $$\text{Span}(I) \ne V$$ does imply that $$I \ne G$$ and also $$I'= I \subseteq V$$?

If $$\text{Span}(I')=V$$ we are done. If not,$$\exists g_2 \in G$$ such that $$g_2 \notin \text{Span}(I')$$ and we get a linearly independent set $$I''= I' \cup \left\{g_1 ,g_2\right\}$$. If $$\text{Span}(I'')=V$$ we are done. If not, we continue. Since $$G$$ is a finite set, this process will come to an end.Finally we will get a linearly independent set $$I'= I \cup \left\{g_1 ,g_2,\ldots,g_r\right\} \subseteq G$$ which spans $$V$$ . This is the basis $$B$$ and obviously $$I \subseteq B \subseteq G$$.

I know that $$G$$ is a spanning set of $$V$$ but why a subset of $$G$$ like $$I'= I \cup \left\{g_1 ,g_2,\ldots,g_r\right\}$$ is also a spanning subset of $$G$$?

• $\text{Span}(I)=\text{Span}(B)$ doesn't mean $I=B$, but it means $I$ is a basis as I is LI and B is a basis. Since we have to show existence of a basis, so the author is saying take $B=I$ Commented Nov 30, 2020 at 7:23

1. We don't know that $$B=I$$; we define $$B$$ as being equal to $$I$$.
2. Since $$\operatorname{Span}(G)=V$$, if $$\operatorname{Span}(I)\ne V$$, then $$I\ne G$$. So, $$I\varsubsetneq G$$. And $$I'\subseteq G$$ because $$I'=I\cup\{g_1\}$$, where $$I$$ is a subset of $$G$$ and $$g_1\in G$$.
3. Because otherwise $$I'\varsubsetneq G$$ and then we could take an element $$g$$ of $$G$$ and add it to $$I'$$ so that $$I'\cup\{g\}$$ would still be linearly independent. But we cannot keep on doing that forever, since $$G$$ is finite. So, sooner or later, that process must come to an end.
• If we define $B=I$ then clearly $I=B \subseteq I$ and from the assumption $I \subseteq G$,besides $I$ is a linearly independent set and so is $B$,but to show that $B$ is a basis we need to show that $I$ spans $V$ ,but I don't see how $I$ spans $V$ and so cannot conclude that $B$ is a basis,that's the problem I have.
– user852833
Commented Nov 30, 2020 at 7:47
• My guess is that that's about my first sentence. Note that there we are assuming that $I$ spans $V$ (it says “If $\operatorname{Span}(I)=V$”) and so nothing is missing here. Commented Nov 30, 2020 at 7:49
• @ José Carlos Santos,you are completely rgith,besides In the second part it looks that you are using the contrapositive of "If $A=B \implies \text{Span}(A)=\text{Span}(B)$".
– user852833
Commented Nov 30, 2020 at 7:57
• @ José Carlos Santos,Is $I \ne G$ helpful to conclude that $G \setminus \text{Span}(I) \ne \emptyset$?
– user852833
Commented Nov 30, 2020 at 8:10
• Indeed I think if $G \setminus \text{span}(I)=\emptyset$ then $\color{blue}{G \subseteq \text{span}(I)}$ ,Moreover from $I \subsetneq G$ it follows that $\color{blue}{\text{span}(I) \subset \text{span}(G)}$ and so $$G \subseteq\text{span}(I) \subset \text{span}(G)=V$$ Which in turn follows that $G \subset V$,and I "guess" this is not possible.
– user852833
Commented Nov 30, 2020 at 9:08