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Rowa
Group Title
how can i know this subset is a subspace:
{(x,y,z) E(element) R^3(real number) : z=x+y}
 one year ago
 one year ago
Rowa Group Title
how can i know this subset is a subspace: {(x,y,z) E(element) R^3(real number) : z=x+y}
 one year ago
 one year ago

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KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
Let \(S=\{(x,y,z)\in\mathbb{R}^3\;\;z=x+y\}\). To prove something is a subspace, you need to satisfy three different conditions. 1. \(\vec0\in S\). 2. If \(\vec{v},\vec{u}\in S\), then \(\vec{v}+\vec{u}\in S\). 3. If \(\vec{u}\in S\), and \(c\in\mathbb{R}\), then \(c\vec{u}\in S\).
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
The first condition is straightforward. \(\vec0=(0,0,0)\), and \(0=0+0\), so this is good. Can you prove the second condition yourself?
 one year ago

Rowa Group TitleBest ResponseYou've already chosen the best response.0
so..., x+y=z, and z is an element of R^3..
 one year ago

Rowa Group TitleBest ResponseYou've already chosen the best response.0
somehow i know thats wrong..
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
Not quite. \((x,y,z)\) is an element of \(\mathbb{R}^3\) such that \(x+y=z\), and \(z\) is an element of \(\mathbb{R}\).
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
And if you have two vectors \((x_0,y_0,z_0)\) and \((x_1,y_1,z_1)\) in \(S\), then \[(x_0,y_0,z_0)+(x_1,y_1,z_1)=(x_0+x_1,y_0+y_1,z_0+z_1)\]Using this can you try to do the second condition?
 one year ago

Rowa Group TitleBest ResponseYou've already chosen the best response.0
i don't see any problem with the second condition..
 one year ago

Rowa Group TitleBest ResponseYou've already chosen the best response.0
If there's only 1 vector (x,y,z), how can i add to see if it follows the second condition *confused*
 one year ago

Rowa Group TitleBest ResponseYou've already chosen the best response.0
ok, i'm slowly getting it
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
You need to take two vectors. So let's use the vectors I wrote out above. Let \(\vec{v}=(x_0,y_0,z_0)\) and let \(\vec{u}=(x_1,y_1,z_1)\). Then we need to check if \[(x_0,y_0,z_0)+(x_1,y_1,z_1)=(x_0+x_1,y_0+y_1,z_0+z_1)\]is still in \(S\). To check this, we just need to check if \[x_0+x_1+y_0+y_1=z_0+z_1.\]However, since \(x_0+y_0=z_0\) and \(x_1+y_1=z_1\), (this is true since both \(\vec{v}\) and \(\vec{u}\) are in \(S\), so they have that property), so it's true, and \(\vec{v}+\vec{u}\in S\).
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
That make sense?
 one year ago

Rowa Group TitleBest ResponseYou've already chosen the best response.0
yea, i get it
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
Then, can you attempt to the third condition?
 one year ago

Rowa Group TitleBest ResponseYou've already chosen the best response.0
so..: c(x,y,z)= cx+cy+cz, cz=cy+cx ??? something like this?
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
That looks pretty good. You just need to make sure that you've proven that cz=cy+cx.
 one year ago

Rowa Group TitleBest ResponseYou've already chosen the best response.0
z=y+x => cz=cy+cx ?
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
Basically. I would add the step that \(z=x+y\implies cz=c(x+y)\implies cz=cx+cy\). But otherwise, that's the correct reasoning.
 one year ago

Rowa Group TitleBest ResponseYou've already chosen the best response.0
nice...Thank you so much
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
You're welcome.
 one year ago
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