# Subspace question...

• Jul 14th 2009, 08:46 PM
Danneedshelp
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
• Jul 14th 2009, 09:08 PM
Jose27
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.
• Jul 14th 2009, 09:59 PM
Danneedshelp
Quote:

Originally Posted by Jose27
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