scalar multiplied by a vector proof

**Jskid** · February 9th 2011, 07:21 PM

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.

**TheEmptySet** · February 9th 2011, 07:27 PM

Originally Posted by Jskid

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 $4^3(x+y+z)$

**dwsmith** · February 9th 2011, 08:14 PM

Originally Posted by Jskid

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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