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A componentwise proof

mark

Type out a component-wise proof that the scalar product is distributive over the addition of two-dimensional vectors. That is, for all real numbers r and for all two-dimensional vectors \vec{u} and \vec{v},

r(\vec{u} + \vec{v}) = r\vec{u} + r\vec{v}.

I’ll get you started:

Let \vec{u} = \langle u_1, u_2 \rangle and let \vec{v} = \langle v_1,v_2 \rangle. Then

\vec{u} + \vec{v} = \cdots

Also, don’t forget that you can right-click or cntrl-click on typeset math to see how it was entered in LaTeX. If I do that with the \vec{v} = \langle v_1,v_2 \rangle above, you should see the following:

This doesn’t include the dollar signs that are needed. Thus I typed in:

$\vec{v} = \langle v_1,v_2 \rangle$

audrey

Here’s my proof:

Let \vec{u} = \langle u_1, u_2 \rangle and let \vec{v} = \langle v_1,v_2 \rangle. Then

\begin{aligned} r(\vec{u} + \vec{v}) &= r(\langle u_1, u_2 \rangle + \langle v_1,v_2 \rangle) \\ &= r\langle u_1 + v_1, u_2 + v_2 \rangle \\ &= \langle r(u_1 + v_1), r(u_2 + v_2) \rangle \\ &= \langle ru_1 + rv_1, ru_2 + rv_2 \rangle \\ &= \langle ru_1, ru_2 \rangle + \langle rv_1, rv_2 \rangle \\ &= r\langle u_1, u_2 \rangle + r\langle v_1, v_2 \rangle = r\vec{u} + r\vec{v}. \, \Box \end{aligned}