Linearly independent, Linearly dependent

$\spadesuit$ Cross Product $\spadesuit$

To consider the cross product of two vectors, we are back to the geometric vectors. The definition of the cross product is not as easy as the inner product. But it is very important in applied mathematics. Here we restrict ourselves in 3D applications. The reason is that there is no simple generalization in other vector space.

**Figure 1.4:** Cross Product
$\includegraphics[width=5cm]{LALG/Fig1-4.eps}$

Definition 1..6

Given any geometric vectors A,B, the magnitude is the same as the area of the parallelogram with vector A and B for sides. The direction is right angle to A and B and follows the right hand rule. Then this vector is called the cross product of A and B, and denoted by ${\bf A}\times{\bf B}$ . For if ${\bf A} = {\bf0}, {\bf B} = {\bf0}, \theta = 0$ , there is no vector orthogonal to A and B. But in this case we define ${\bf A}\times{\bf B} = 0$ .

In 3D space, the cross product of vectors $(a_{1},a_{2},a_{3})$ and $(b_{1},b_{2},b_{3})$ is defined as follows:

$\displaystyle (a_{1},a_{2},a_{3})\times(b_{1},b_{2},b_{3}) = (a_{2}b_{3} - a_{3}b_{2}, a_{3}b_{1} - a_{1}b_{3},a_{1}b_{2} - a_{2}b_{1}).$

If we use the determinant(study later), we can express the above by the following:

$\displaystyle (a_{1},a_{2},a_{3})\times(b_{1},b_{2},b_{3}) = \left\vert \begin{... ... j}&{\bf k}\\ a_{1}&a_{2}&a_{3}\\ b_{1}&b_{2}&b_{3} \end{array}\right\vert .$

Example 1..15

Expree $\vert{\bf A}\times{\bf B}\vert$ using the inner product.

Answer

$\displaystyle \vert{\bf A}\times{\bf B}\vert^{2}$	$\displaystyle =$	$\displaystyle \vert{\bf A}\vert^2\vert{\bf B}\vert^2\sin^{2}{\theta}$
	$\displaystyle =$	$\displaystyle \vert{\bf A}\vert^2\vert{\bf B}\vert^2(1 - \cos^2{\theta})$
	$\displaystyle =$	$\displaystyle \vert{\bf A}\vert^2\vert{\bf B}\vert^2 - \vert{\bf A}\vert^2\vert{\bf B}\vert^2\cos^{2}{\theta}$
	$\displaystyle =$	$\displaystyle ({\bf A}\cdot{\bf A})({\bf B}\cdot{\bf B}) - ({\bf A}\cdot{\bf B})^2 .$

よって, $\vert{\bf A}\times{\bf B}\vert = \sqrt{({\bf A}\cdot{\bf A})({\bf B}\cdot{\bf B}) - ({\bf A}\cdot{\bf B})^2}.$ $\blacksquare$

In mechanics, the moment $m$ of the force F about $O$ is given by $m = \vert{\bf F}\vert d$ , where $d$ is the distance from $O$ to the line of action. Suppose that r is a vector connecting $P$ and $O$ . Then $d = \vert{\bf r}\vert\sin{\theta}$ . Thus we can write $m$ as

**Figure 1.5:** Moment of F about
$\includegraphics[width=5.5cm]{LALG/Fig1-5.eps}$

$\displaystyle m = \vert{\bf F}\Vert{\bf r}\vert\sin{\theta} = \vert{\bf F}\times{\bf r}\vert = \vert{\bf r}\times{\bf F}\vert$

In this case, we say ${\bf m} = {\bf r}\times{\bf F}$ is called moment vector of F about

For any geometric vectors A,B,C, $({\bf A}\times{\bf B})\cdot{\bf C}$ is a real number. So, it is called a scalar triple product. ${\bf A}\times({\bf B}\times{\bf C})$ is a vector. So, it is called a vector triple product.

The vector triple product ${\bf A}\times({\bf B}\times{\bf C})$ is a vector on the plane formed by B and C. Then if B and C are not parallel, ${\bf A}\times({\bf B}\times{\bf C})$ can be written as follows:.

$\displaystyle {\bf A}\times({\bf B}\times{\bf C}) = ({\bf A}\cdot{\bf C}){\bf B} - ({\bf A}\cdot{\bf B}){\bf C} .$

$\spadesuit$ Linear Combination $\spadesuit$

In vector sapace, the addition and scalar multiplication are essential. An addition is the operation of two vectors. In the vector space, associative law holds, so, we can add three, four vectors and so on. This operation is called linear combination.

Definition 1..7

For any vector space $V$

and any vectors ${\bf v}_{1},{\bf v}_{2}, {\bf v}_{3}, \ldots , {\bf v}_{n}$ and any real numbers $c_{1},c_{2},c_{3},\ldots,c_{n}$ ,

$\displaystyle c_{1}{\bf v}_{1} + c_{2}{\bf v}_{2} + c_{3}{\bf v}_{3} + \cdots + c_{n}{\bf v}_{n}$

is called a linear combination.

Here we want you to notice that in the vector space $V$ any linear combination of vectors in $V$ is again a vector in $V$ .

Next we consider a linear combination of continuous functions $\{1, \cos^2{x}, \sin^2{x}\}$

$\begin{displaymath}\begin{array}{llll} (a)& 1 + \cos^2{x} + \sin^2{x}&(b)& 1 - 2... ...\\ (c)& -1 + \cos^2{x} + \sin^2{x}&(d)& 4\sin^2{x} \end{array}\end{displaymath}$

For all $c_{i}$ are 0, a linear combination becomes 0. Now we ask a question. If some of $c_{i}$ are not 0, is it possible to have the linear combination be 0. Look at the example $(c)$ . $-1 + \cos^2{x} + \sin^2{x}$ is 0. In this way, if some of $c_{i}$ are not 0, yet the linear combination is 0. We say the set of vectors $\{1, \cos^2{x}, \sin^2{x}\}$ is linearly dependent. If no such $c_{i}$ exits, we say the set of vectors is linearly independent.

Definition 1..8

$\displaystyle c_{1}{\bf v}_{1} + c_{2}{\bf v}_{2} + c_{3}{\bf v}_{3} + \cdots + c_{n}{\bf v}_{n} = {\bf0}$

implies $c_{1} = c_{2} = c_{3} =\cdots = c_{n} = 0$ . Then ${\bf v}_{1}, {\bf v}_{2}, {\bf v}_{3},\cdots , {\bf v}_{n}$ is linearly independent. Otherwise, ${\bf v}_{1}, {\bf v}_{2}, {\bf v}_{3},\cdots , {\bf v}_{n}$ is linearly dependent.

Example 1..16

Show the set of vectors $\{{\bf i}, {\bf j}, {\bf k} \}$ is linearly independent.

Answer Let $c_{1}(1,0,0) + c_{2}(0,1,0) + c_{3}(0,0,1) = (0,0,0)$ . Then $c_{1} = c_{2} = c_{3} = 0$ . Thus $\{{\bf i}, {\bf j}, {\bf k} \}$ is linearly independent.. $\blacksquare$

Example 1..17

Show the set of vectors $\{3{\bf i} + 2{\bf j}, {\bf i} + 5{\bf k}, 6{\bf i} + {\bf j} \}$ is linearly independent..

Answer A linear combination of $3{\bf i} + 2{\bf j}, {\bf i} + 5{\bf k}, 6{\bf i} + {\bf j}$ is $c_{1}(3{\bf i} + 2{\bf j}) + c_{2}({\bf i} + 5{\bf k}) + c_{3}(6{\bf i} + {\bf j})$ . Set this to 0, we have

$\displaystyle (3c_{1} + c_{2} + 6c_{3}){\bf i} + (2c_{1} + c_{3}) {\bf j} + 5c_{2}{\bf k} = {\bf0} .$

Note that the set of $\{{\bf i}, {\bf j}, {\bf k} \}$ is linearly independent, the coefficeint of $\{{\bf i}, {\bf j}, {\bf k} \}$ are 0. Thus, we have the following system of equations.

$\displaystyle \left(\begin{array}{rrrl} 3c_{1} & + c_{2} & + 6c_{3} & = 0\\ 2c_{1} & & + c_{3} & = 0\\ & 5c_{2} & & = 0 . \end{array}\right .$

Solving this system of equations, we have $c_{1} = c_{2} = c_{3} = 0$ . Thus, the set of vectors is linearly independent. $\blacksquare$

Example 1..18

Determine whether $\{ {\bf i} + {\bf j} + 3{\bf k}, {\bf i} + 2{\bf k}, {\bf i} - 2{\bf j} \}$ is linearly independent or dependent..

Answer A linear combintion of $\{ {\bf i} + {\bf j} + 3{\bf k}, {\bf i} + 2{\bf k}, {\bf i} - 2{\bf j} \}$ is set to 0. Then we have

$\displaystyle c_{1}({\bf i} + {\bf j} + 3{\bf k}) + c_{2}({\bf i} + 2{\bf k}) + c_{3}({\bf i} - 2{\bf j}) = {\bf0}.$

Solving this equation, we have

$\displaystyle c_{1} +c_{2} +c_{3} = 0, c_{1} - 2c_{3} = 0, 3c_{1} + 2c_{2} = 0$

This equation has for example solutions like $c_{1} = 2, c_{2} = -3, c_{3} = 1$ , which is not 0. Therefore, $\{ {\bf i} + {\bf j} + 3{\bf k}, {\bf i} + 2{\bf k}, {\bf i} - 2{\bf j} \}$ is linearly dependent.. $\blacksquare$

If you look at the above example more carefully, we can express ${\bf i} - 2{\bf j} = -2({\bf i} + {\bf j} + 3{\bf k}) + 3({\bf i} + 2{\bf k})$ . In other words, one of the vector is a linear combination of others.

Theorem 1..3

If a vector w is a linear combination of vectors ${\bf v}_{1}, {\bf v}_{2}, \ldots , {\bf v}_{n}$ , then w and ${\bf v}_{1}, {\bf v}_{2}, \ldots , {\bf v}_{n}$ are linearly dependent..

Conversely, if some vectors are linearly dependent each other, then one of them can be expressed by a linear combination of other vectors.

Proof Suppose that

$\displaystyle {\bf w} = c_{1}{\bf v}_{1} + c_{2}{\bf v}_{2} + \cdots + c_{n}{\bf v}_{n}$

Then

$\displaystyle c_{1}{\bf v}_{1} + c_{2}{\bf v}_{2} + \cdots + c_{n}{\bf v}_{n} - 1{\bf w} = {\bf0}$

and ${\bf v}_{1}, {\bf v}_{2}, \ldots , {\bf v}_{n}, {\bf w}$ is linearly dependent.

Conversely, if the set of ${\bf v}_{1}, {\bf v}_{2}, \ldots , {\bf v}_{n}$ is linearly dependent, then

$\displaystyle c_{1}{\bf v}_{1} + c_{2}{\bf v}_{2} + \ldots + c_{n}{\bf v}_{n} = {\bf0}$

and at least one of coefficents $c_{1},c_{2}, \ldots , c_{n}$ is not 0. Then we can let one of them be $c_{n} \neq 0$ . Then ${\bf v}_{n}$ can be written as

$\displaystyle {\bf v}_{n} = -\frac{1}{c_{n}}(c_{1}{\bf v}_{1} + c_{2}{\bf v}_{2} + \ldots + c_{n-1}{\bf v}_{n-1})$

Therefore, ${\bf v}_{n}$ is a linear combination of vectors ${\bf v}_{1}, {\bf v}_{2}, \ldots , {\bf v}_{n-1}$ . $\blacksquare$

To check to see whether the set of given geometric vectors is independent or not, the scalar triple product is usefull.

Theorem 1..4

Show that the set of geometric vectors A,B,C is linearly independent if and only if the scalar triple product ${\bf A}\cdot({\bf B}\times{\bf C}) \neq 0$ .

Proof By the exercise 1.3, the scalar triple product represents the volume of parallelpiped. Then ${\bf A}\cdot({\bf B}\times{\bf C}) = 0$ implies that A,B,C are on the same plane. Also, by Theorem1.1, the set of vectors A,B,C is linearly dependent. $\blacksquare$

Example 1..19

Show that the set of vectors ${\bf A} = {\bf i} + 2{\bf j} - {\bf k}, {\bf B} = 2{\bf i} - {\bf j} - {\bf k}, {\bf C} = -{\bf i} + 3{\bf j} + 4{\bf k}$ is linearly independet.

Answer ${\bf A}\cdot({\bf B}\times{\bf C}) = ({\bf i} + 2{\bf j} - {\bf k})\cdot(-{\bf i} - 7{\bf j} + 5{\bf k}) = -20$ . Thus, it is linearly independent. $\blacksquare$

Exercise1-6

1. For vectors ${\bf A} = (1,2,-3), {\bf B} = (2,-1,1), {\bf C} = (4,2,2)$ , find the followings:

(a) ${\bf A}\times{\bf B}$ (b) ${\bf C}\times({\bf A}\times{\bf B})$ (c) ${\bf C}\cdot({\bf A}\times{\bf B})$

2. Find the equation of the plane going thru a point $(1,0,1)$ and parallel to the plane with the vector ${\bf i} + {\bf j} - {\bf k}$ and $2{\bf i} + 3{\bf j} + 2{\bf k}$ for sides.

3. Find the equation of the plane going thru $(2,0,-1), (3,2,1)$ and perpendicular to the plane $x - 2y + 3z - 4 = 0$ .

4. Find the area of the triangle whose sides are given by ${\bf A} = {\bf i} + 3{\bf j} - {\bf k}, {\bf B} = 2{\bf i} + {\bf j} + {\bf k}$ ..

5. Find the moment vector of ${\bf F} = {\bf i} + 3{\bf j} + {\bf k}$ around the point $(2,-1,1)$

6. The volume of the parallelogram composed by the vectors A,B,C is the same as the absolute value of

$\displaystyle {\bf A}\cdot({\bf B}\times{\bf C})$

7. Given ${\bf A} = 2{\bf e}_{1} + 5{\bf e}_{2} - {\bf e}_{3}, {\bf B} = {\bf e}_{1} - 2{\bf e}_{2} - 4{\bf e}_{3}$ and ${\bf e}_{1}\times{\bf e}_{2} = {\bf i} - {\bf j}, {\bf e}_{1}\times{\bf e}_{3} = {\bf j} + {\bf k},{\bf e}_{2}\times{\bf e}_{3} = {\bf i} + {\bf k}$ . Find ${\bf A}\times{\bf B}$ .

8. Determine whether $\{4{\bf i} - 3{\bf j} + {\bf k}, 10{\bf i} - 3{\bf j}, 2{\bf i} -6{\bf j} + 3{\bf k}\}$ is linearly independent or not.

9. Show the following functions are linearly independent on any interval $(a,b)$

(a) $\{1, x, x^2\}$ (b) $\{\sin{x}, \cos{x}\}$

10. Show that geometric vectors A, B is linearly independent if and only if ${\bf A} \times {\bf B} \neq {\bf0}$