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

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KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.1Let \(S=\{(x,y,z)\in\mathbb{R}^3\;\;z=x+y\}\). 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 \(c\vec{u}\in S\).

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

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

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

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

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.1And 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?

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

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

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

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.1You 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 \(\vec{v}+\vec{u}\in S\).

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

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

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

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.1Basically. 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.
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