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Math Help - scalar multiplied by a vector proof

  1. #1
    Member Jskid's Avatar
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    scalar multiplied by a vector proof

    Show that if c is a scalar then ||cu||=|c|||u||

    let u =(x_1, x_2, ..., x_n) then ||u||= \sqrt{x^2_1+x^2_2+...+x^2_n}
    cu= (cx_1, cx_2,...,cx_n) then ||cu||= \sqrt{(cx_1)^2+(cx_2)^2+...+(cx_n)^2}=\sqrt{c^2x_1  ^2+c^2x_2^2+...+c^2x_n^2}=\sqrt{c^{2n}(x^2_1+x^2_2  +...+x^2_n)}
    I guess I messed something up because this looks like I disproved it.
    Last edited by Jskid; February 9th 2011 at 07:22 PM. Reason: changed latex
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by Jskid View Post
    Show that if c is a scalar then ||cu||=|c|||u||

    let u =(x_1, x_2, ..., x_n) then ||u||= \sqrt{x^2_1+x^2_2+...+x^2_n}
    cu= (cx_1, cx_2,...,cx_n) then ||cu||= \sqrt{(cx_1)^2+(cx_2)^2+...+(cx_n)^2}=\sqrt{c^2x_1  ^2+c^2x_2^2+...+c^2x_n^2}=\sqrt{c^{2n}(x^2_1+x^2_2  +...+x^2_n)}
    I guess I messed something up because this looks like I disproved it.
    Where did the n come from in the c^{2n}

    If I factor 4x+4y+4z I get 4(x+y+z) not 4^3(x+y+z)
    Last edited by TheEmptySet; February 14th 2011 at 12:47 PM. Reason: typo
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  3. #3
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    Quote Originally Posted by Jskid View Post
    Show that if c is a scalar then ||cu||=|c|||u||

    let u =(x_1, x_2, ..., x_n) then ||u||= \sqrt{x^2_1+x^2_2+...+x^2_n}
    cu= (cx_1, cx_2,...,cx_n) then ||cu||= \sqrt{(cx_1)^2+(cx_2)^2+...+(cx_n)^2}=\sqrt{c^2x_1  ^2+c^2x_2^2+...+c^2x_n^2}=\sqrt{c^{2n}(x^2_1+x^2_2  +...+x^2_n)}
    I guess I messed something up because this looks like I disproved it.
    \sqrt{c^{2}(x^2_1+x^2_2+...+x^2_n)}=\sqrt{c^2}\cdo  t\sqrt{(x^2_1+x^2_2+...+x^2_n)}=\cdots
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