anonymous
  • anonymous
how can i know this subset is a subspace: {(x,y,z) E(element) R^3(real number) : z=x+y}
Linear Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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KingGeorge
  • KingGeorge
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 \(c\vec{u}\in S\).
KingGeorge
  • KingGeorge
The first condition is straightforward. \(\vec0=(0,0,0)\), and \(0=0+0\), so this is good. Can you prove the second condition yourself?
anonymous
  • anonymous
so..., x+y=z, and z is an element of R^3..

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anonymous
  • anonymous
somehow i know thats wrong..
KingGeorge
  • KingGeorge
Not quite. \((x,y,z)\) is an element of \(\mathbb{R}^3\) such that \(x+y=z\), and \(z\) is an element of \(\mathbb{R}\).
anonymous
  • anonymous
oooh, ok
KingGeorge
  • KingGeorge
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?
anonymous
  • anonymous
k
anonymous
  • anonymous
i don't see any problem with the second condition..
anonymous
  • anonymous
If there's only 1 vector (x,y,z), how can i add to see if it follows the second condition *confused*
anonymous
  • anonymous
ok, i'm slowly getting it
KingGeorge
  • KingGeorge
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 \(\vec{v}+\vec{u}\in S\).
KingGeorge
  • KingGeorge
That make sense?
anonymous
  • anonymous
yea, i get it
KingGeorge
  • KingGeorge
Then, can you attempt to the third condition?
anonymous
  • anonymous
so..: c(x,y,z)= cx+cy+cz, cz=cy+cx ??? something like this?
KingGeorge
  • KingGeorge
That looks pretty good. You just need to make sure that you've proven that cz=cy+cx.
anonymous
  • anonymous
z=y+x => cz=cy+cx ?
KingGeorge
  • KingGeorge
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.
anonymous
  • anonymous
nice...Thank you so much
KingGeorge
  • KingGeorge
You're welcome.

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