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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

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  1. KingGeorge Group Title
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    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
  2. KingGeorge Group Title
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    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
  3. Rowa Group Title
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    so..., x+y=z, and z is an element of R^3..

    • one year ago
  4. Rowa Group Title
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    somehow i know thats wrong..

    • one year ago
  5. KingGeorge Group Title
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    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
  6. Rowa Group Title
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    oooh, ok

    • one year ago
  7. KingGeorge Group Title
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    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
  8. Rowa Group Title
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    k

    • one year ago
  9. Rowa Group Title
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    i don't see any problem with the second condition..

    • one year ago
  10. Rowa Group Title
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    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
  11. Rowa Group Title
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    ok, i'm slowly getting it

    • one year ago
  12. KingGeorge Group Title
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    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
  13. KingGeorge Group Title
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    That make sense?

    • one year ago
  14. Rowa Group Title
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    yea, i get it

    • one year ago
  15. KingGeorge Group Title
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    Then, can you attempt to the third condition?

    • one year ago
  16. Rowa Group Title
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    so..: c(x,y,z)= cx+cy+cz, cz=cy+cx ??? something like this?

    • one year ago
  17. KingGeorge Group Title
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    That looks pretty good. You just need to make sure that you've proven that cz=cy+cx.

    • one year ago
  18. Rowa Group Title
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    z=y+x => cz=cy+cx ?

    • one year ago
  19. KingGeorge Group Title
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    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
  20. Rowa Group Title
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    nice...Thank you so much

    • one year ago
  21. KingGeorge Group Title
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    You're welcome.

    • one year ago
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