# Thread: Linear algebra : base, vector space, prove that.

1. ## Linear algebra : base, vector space, prove that.

Hi MHF,
I don't know how to prove what I must prove, but I've done something.
Let $\displaystyle \bold{B}$ be a base of the vector space $\displaystyle V$ which has a finite dimension. Let $\displaystyle S \subseteq V$ such that $\displaystyle \bold{B} \subseteq \text{span } (S)$. Prove that $\displaystyle \text{span } (S)=V$.
My attempt : We have that $\displaystyle \dim S \leq \dim V$. And also that $\displaystyle \dim \bold{B} \leq \dim S$, but as $\displaystyle \dim \bold{B}=\dim V$ we have $\displaystyle \dim S=\dim V$. So the conclusion follows. Am I right?

2. Originally Posted by arbolis
Hi MHF,
I don't know how to prove what I must prove, but I've done something.
Let $\displaystyle \bold{B}$ be a base of the vector space $\displaystyle V$ which has a finite dimension. Let $\displaystyle S \subseteq V$ such that $\displaystyle \bold{B} \subseteq \text{span } (S)$. Prove that $\displaystyle \text{span } (S)=V$.
My attempt : We have that $\displaystyle \dim S \leq \dim V$. And also that $\displaystyle \dim \bold{B} \leq \dim S$, but as $\displaystyle \dim \bold{B}=\dim V$ we have $\displaystyle \dim S=\dim V$. So the conclusion follows. Am I right?
hmmm,

here is some nice way..
note that every element of $\displaystyle V$ can be written as a linear combination of the elements in $\displaystyle B$.. but since $\displaystyle B \subseteq \mbox{span }{S}$, then every element in $\displaystyle B$ can be written as a linear combination of the elements in $\displaystyle S$... thus every element in $\displaystyle V$ can be written as a linear combination of the elements in $\displaystyle S$.

3. We suppose $\displaystyle S=\{ \vec{v}_1 \cdots \vec{v}_m \cdots \vec{v}_n \}, \mathfrak{B}=\{ \vec{v}_1 \cdots \vec{v}_m \}, \mathfrak{B}\subseteq span(S)$
$\displaystyle \because V=span(\mathfrak{B})\ and\ S \subseteq V \therefore S=span(\mathfrak{B}), \{ \vec{v}_{m+1} \cdots \vec{v}_n \}\$ $\displaystyle are\ linearly\ dependent\ elements\therefore span(S)=span(\mathfrak{B})=V$

4. Originally Posted by math2009
We suppose $\displaystyle S=\{ \vec{v}_1 \cdots \vec{v}_m \cdots \vec{v}_n \}, \mathfrak{B}=\{ \vec{v}_1 \cdots \vec{v}_m \}, \mathfrak{B}\subseteq span(S)$
$\displaystyle \because V=span(\mathfrak{B})\ and\ S \subseteq V \therefore S=span(\mathfrak{B}), \{ \vec{v}_{m+1} \cdots \vec{v}_n \}\ are\ linearly\ dependent$

and what do you want to show with this?

5. I didn't finish writing, please refresh web page.

6. actually, there is no need to say that the the latter set is a lin. dep set.. as long as you can show that span B = span S, your done..