Results 1 to 6 of 6

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

  1. #1
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor kalagota's Avatar
    Joined
    Oct 2007
    From
    Taguig City, Philippines
    Posts
    1,026
    Quote Originally Posted by arbolis View Post
    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 $\displaystyle S$.
    Last edited by kalagota; Feb 20th 2009 at 06:17 PM. Reason: gave some proof..
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jan 2009
    Posts
    153
    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 \}\ $ $\displaystyle are\ linearly\ dependent\ elements\therefore span(S)=span(\mathfrak{B})=V$
    Last edited by math2009; Feb 20th 2009 at 08:58 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor kalagota's Avatar
    Joined
    Oct 2007
    From
    Taguig City, Philippines
    Posts
    1,026
    Quote Originally Posted by math2009 View Post
    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?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jan 2009
    Posts
    153
    I didn't finish writing, please refresh web page.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor kalagota's Avatar
    Joined
    Oct 2007
    From
    Taguig City, Philippines
    Posts
    1,026
    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..
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. prove that a set V is a vector space
    Posted in the Advanced Algebra Forum
    Replies: 20
    Last Post: Dec 22nd 2011, 02:51 PM
  2. Prove that the set is not a subspace of vector space
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Mar 17th 2010, 06:04 PM
  3. Linear Algebra: Vector Space!
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Jan 31st 2009, 10:48 AM
  4. [Linear Algebra] problem with finding base and dimension
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Jan 8th 2009, 05:40 PM
  5. vector space prove!
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Aug 18th 2008, 09:29 PM

Search Tags


/mathhelpforum @mathhelpforum