5. Vectors

c. Scalar Multiplication

3. Properties

We here list the algebraic properties of scalar multiplication and its relation to vector addition and magnitude.

In the following, \(a\) and \(b\) are scalars (real numbers) while \(\vec u\), \(\vec v\), and \(\vec w\) are vectors. Also, \(\vec 0\) is the vector whose components are all zero and \(-\vec v\) denotes the vector whose components are the negatives of those of \(\vec v\).

First, there are two properties involving just scalar multiplication.

Let \(a\) and \(b\) be arbitrary scalars and \(\vec v\) be an arbitrary vector. Then,

  1. Scalar Multiplication is Associative:   \((ab)\vec v=a(b\vec v)\)  

\[\begin{aligned} (ab)\vec v &=\left\langle abv_1,abv_2\right\rangle \\ &=a\left\langle bv_1,bv_2\right\rangle =a(b\vec v) \end{aligned}\]

  1. \(1\) is the Multiplicative Identity:   \(1\,\vec v=\vec v\)  

\[ 1\,\vec v=\left\langle1v_1,1v_2\right\rangle =\left\langle v_1,v_2\right\rangle=\vec v \]

Second, there are six properties relating vector addition and scalar multiplication.

Let \(a\) and \(b\) be arbitrary scalars and \(\vec u\) and \(\vec v\) be arbitrary vectors. Then,

  1. Scalar Multiplication Distributes over Ordinary Addition:   \((a+b)\vec v=a\vec v+b\vec v\)  

\[\begin{aligned} (a+b) \vec v &=\left\langle(a+b)v_1,(a+b)v_2\right\rangle \\ &=\left\langle av_1+bv_1,av_2+bv_2\right\rangle \\ &=\left\langle av_1,av_2\right\rangle+\left\langle bv_1,bv_2\right\rangle \\ &=a\vec v+b\vec v \end{aligned}\]

  1. Scalar Multiplication Distributes over Vector Addition:   \(a(\vec u+\vec v)=a\vec u+a\vec v\)  

\[\begin{aligned} a(\vec u+\vec v) &=a\left\langle u_1+v_1,u_2+v_2\right\rangle \\ &=\left\langle a(u_1+v_1),a(u_2+v_2)\right\rangle \\ &=\left\langle au_1+av_1,au_2+av_2\right\rangle \\ &=\left\langle au_1,au_2\right\rangle+\left\langle av_1,av_2\right\rangle \\ &=a\vec u+a\vec v \end{aligned}\]

  1. \((-1)\vec v\) is the Negative of \(\vec v\):   \((-1)\vec v=-\vec v\)  

\[ (-1)\,\vec v=\left\langle-1v_1,-1v_2\right\rangle=-\vec v \]

  1. \(0\) times any Vector is \(\vec 0\):   \(0\,\vec v=\vec0\)  

\[ 0\,\vec v=\left\langle0v_1,0v_2\right\rangle =\left\langle 0,0\right\rangle=\vec0 \]

  1. Any Scalar times \(\vec0\) is \(\vec0\):   \(a\vec0=\vec0\)  

\[ a\vec0=\left\langle a0,a0\right\rangle =\left\langle0,0\right\rangle=\vec0 \]

  1. Zero Product: If \(a\vec v=\vec0\), then either \(a=0\) or \(\vec v=\vec0\):  

Either \(a=0\) or \(a\ne0\). If \(a\ne0\), multiply both sides of \(a\vec v=\vec0\) by \(\dfrac{1}{a}\): \[ \dfrac{1}{a}(a\vec v)=\dfrac{1}{a}\vec0 \] On the left, use (1) associativity and (2) \(1\) is the unit: \[ \dfrac{1}{a}(a\vec v)=\left(\dfrac{1}{a}a\right)\vec v=1\vec v=\vec v \] On the right, use (7) multiple of \(\vec0\): \[ \dfrac{1}{a}\vec0=\vec0 \] Together, these say: \[ \vec v=\vec0 \]

Here is an alternate proof using components. \[\begin{aligned} &&a\vec v&=\vec0 \\ &\implies&\left\langle av_1,av_2\right\rangle &=\left\langle0,0\right\rangle \\ &\implies&av_1=0, &\quad av_2=0 \\ \end{aligned}\] So either \(a=0\) or \(v_1=v_2=0\). In the latter case, \(\vec v=\vec0\).

Finally, we have the relation between scalar multiplication and magnitude.

Let \(a\) be an arbitrary scalar and \(\vec v\) be an arbitrary vector. Then,

  1. Scalar Multiplication stretches or shrinks a vector:   \(|a\,\vec v|=|a|\,|\vec v|\)  

\[\begin{aligned} |a\vec v| &=\sqrt{\left(a\,v_1\right)^2+\left(av_2\right)^2} =\sqrt{a^2\left({v_1}^2+{v_2}^2\right)} \\ &=\sqrt{a^2}\sqrt{ {v_1}^2+{v_2}^2} =|a|\,|\vec v| \end{aligned}\]

PY: Add disacussion of abstract proofs of Props 5-8

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