Results 1 to 3 of 3

Math Help - Subspace question...

  1. #1
    Senior Member Danneedshelp's Avatar
    Joined
    Apr 2009
    Posts
    303

    Subspace question...

    Hey there,

    I have been studying all day for an exam I have tomorrow and now I find myself stuck on something I really ought to know...

    Here is the problem: Let \vec{x}, \vec{y}, and \vec{z} be vectors in a vector space V. Show that the set of all linear combinations of \vec{x}, \vec{y}, \vec{z}

    W=\{a\vec{x}+b\vec{y}+c\vec{z} : a,b,c\in{\mathbb{R}}\}

    is a subspace of V.

    A: Well, this is clearly the span of \vec{x,y,z} and therefore is a subspace of V. Even so, the teacher wants me to show W is nonempty, closed under addition, and closed under scalar multiplication.

    So, would I just choose some r,s\in{\mathbb{R}} and (\vec{ax},\vec{by},\vec{cz}),(\vec{ax'},\vec{by'},  \vec{cz'})\in{W} and then show W is closed under addition and scalar multiplication by showing r\vec{v}+s\vec{v'}\in{W}?

    Do the constants in W have to be primed or am I just concerned with distinguishing between my vectors?

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Apr 2009
    From
    México
    Posts
    721
    What is (\vec{ax},\vec{by},\vec{cz}),(\vec{ax'},\vec{by'},  \vec{cz'})\in{W}?

    To show that W is closed under addition and scalar multiplication take w_1=a_1x + b_1y +c_1z and w_2=a_2x+b_2y+c_2z where a_i,b_i \in \mathbb{R} and r \in \mathbb{R}. Now w_1+w_2=(a_1x + b_1y +c_1z)+(a_2x+b_2y+c_2z)=(a_1+a_2)x+(b_1+b_2)y+(c_1  +c_2)z and since \mathbb{R} is closed under addition we get that w_1+w_2=a_3x+b_3y+c_3z where a_3=a_1+a_2 and so on. In the same manner rw_1=r(a_1x+b_1y+c_1z)=(ra_1)x+(rb_1)y+(rc_1)z and \mathbb{R} is closed under multiplication...and the rest you can do.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member Danneedshelp's Avatar
    Joined
    Apr 2009
    Posts
    303
    Quote Originally Posted by Jose27 View Post
    What is (\vec{ax},\vec{by},\vec{cz}),(\vec{ax'},\vec{by'},  \vec{cz'})\in{W}?

    To show that W is closed under addition and scalar multiplication take w_1=a_1x + b_1y +c_1z and w_2=a_2x+b_2y+c_2z where a_i,b_i \in \mathbb{R} and r \in \mathbb{R}. Now w_1+w_2=(a_1x + b_1y +c_1z)+(a_2x+b_2y+c_2z)=(a_1+a_2)x+(b_1+b_2)y+(c_1  +c_2)z and since \mathbb{R} is closed under addition we get that w_1+w_2=a_3x+b_3y+c_3z where a_3=a_1+a_2 and so on. In the same manner rw_1=r(a_1x+b_1y+c_1z)=(ra_1)x+(rb_1)y+(rc_1)z and \mathbb{R} is closed under multiplication...and the rest you can do.
    Thank you! Thats how I orginally did it when it was homework last week ha. I saw someone do it another way online and then I got all confused. Thats what I was trying to replicate. I better get some sleep.

    Thanks again
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. R^4 subspace question
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: June 24th 2011, 11:57 PM
  2. Replies: 7
    Last Post: May 17th 2011, 01:31 PM
  3. question about subspace
    Posted in the Advanced Algebra Forum
    Replies: 8
    Last Post: March 1st 2010, 09:25 PM
  4. Subspace question
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: July 11th 2009, 01:15 PM
  5. Vector subspace question
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: May 13th 2008, 12:23 PM

Search Tags


/mathhelpforum @mathhelpforum