Math 511: Linear Algebra

4.1 Vectors in $\mathbb{R}^n$

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4.1.1 Vectors in the Plane

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A vector on the plane has two quantities associated with it; it's magnitude and direction.

The magnitude of a vector in $\mathbb{R}^2$ is the length of the vector, and the direction of such a vector is the angle the vector makes with a vector parallel to $\ihat$.

Figure 1

Notice that this definition does not include where the vector is located on the plane.

Figure 2

All of the vectors in figure 2 are the same vector! In some applications one says that the vector is applied at a specific point $P = (x_0,y_0,z_0)$ if the base of the vector is located at $P$. However the same vector can be applied to any point on the plane.

The coordinates of a vector on the plane are the coordinates of the tip of the vector when the base of the vector is the origin, $(0,0)$. When the base of a vector is the origin, the vector is said to be in standard position. In figure 1, both vectors $\mathbf{x}$ and $\mathbf{y}$ are in standard position.

The standard coordinates of a vector are the coordinates of the tip of the vector in standard position. Often we denotes the coordinates of a vector using the same lower case letter in normal font to indicate it is a scalar, and with a subscript to indicate which coordinate it is.

$$ \mathbf{v} = \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \langle v_1, v_2 \rangle $$

The coordinates are also called the scalar components of the vector since the linear combination of the standard basis vectors, $\ihat$ and $\jhat$ are given by

$$ \mathbf{v} = v_1\ihat + v_2\jhat = v_1\mathbf{e}_1 + v_2\mathbf{e}_2 $$
We will soon be discussing vectors in $\mathbb{R}^n$ so we need to start using $\mathbf{e}_1$, $\mathbf{e}_2$, $\mathbf{e}_3$, $\dots$, $\mathbf{e}_n$ for our standard basis vectors instead of $\ihat$, $\jhat$, and $\khat$.

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4.1.2 The Physicist's Perspective

Vectors in $\mathbb{R}^2$ are most easily thought of using the physicist picture of an arrow representing a magnitude and direction. We also know that we can relate this picture to an ordered list of numbers, the programmer picture. Suppose for now we have a vector $\mathbf{x}\in\mathbb{R}^2$. The vector $\mathbf{x}$ may be considered as both an arrow from the origin to a point $(x_1,x_2)$ and the list of numbers

$$\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}. $$
If $x_1 = 3$ and $x_2 = -2$, then graphically we have

To find the magnitude of a vector $\|\mathbf{x}\|$, we compute the Euclidean distance $d$ from the base to the tip of the vector.

$$ \|\mathbf{x}\| := d = \sqrt{x_1^2 + x_2^2}. $$
This formula should remind you of the Pythagorean theorem.
The angle between two vectors $\mathbf{x}$ and $\mathbf{y}$ is given using dot product and the formula

$$ \cos\theta = \dfrac{\mathbf{x}\cdot\mathbf{y}}{\|\mathbf{x}\|\|\mathbf{y}\|} $$

Exercise 1

If $\mathbf{x} = \begin{bmatrix} 1 \\ 3 \end{bmatrix}$ and $\mathbf{y} = \begin{bmatrix} 0 \\ 2 \end{bmatrix}$, what is $\|\mathbf{x}\|$, $\|\mathbf{y}\|$, and the angle between these two vectors in $\mathbb{R}^2$?

View Solution

$$ \begin{align*} \|\mathbf{x}\| &= \sqrt{ 1^2 + 3^2 } = \sqrt{1 + 9 } =\sqrt{10} \\ \\ \|\mathbf{y}\| &= \sqrt{ 0^2 + 2^2 } = \sqrt{ 0 + 4 } = \sqrt{4} = 2 \\ \\ \cos\left(\theta\right) &= \dfrac{\mathbf{x}\cdot\mathbf{y}}{\|\mathbf{x}\|\|\mathbf{y}\|} = \dfrac{ 1(0) + 3(2) }{ \sqrt{10}\cdot 2 } = \frac{3}{\sqrt{10}} \\ \\ \theta &= \cos^{-1}\left(\frac{3}{\sqrt{10}}\right) \end{align*} $$

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4.1.3 Scalar Multiplication and Vector Addition

The most important concept in linear algebra is the idea of a linear combination. Linear combinations and the properties of vector spaces arise from scalar multiplication and vector addition.

Scalar Multiplication

Scalar multiplication quite literally scales a vector's magnitude by a constant. Most times, these scalars are real numbers, but in some vector spaces they are complex numbers. You are probably already familiar with the programmer's perspective in $\mathbb{R}^2$. Let $\alpha$ be some real number, then

$$ \mathbf{x} = \alpha\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} \alpha x_1 \\ \alpha x_2 \end{bmatrix}. $$
If $\alpha \gt 0$, then $\alpha\mathbf{x}$ is a vector whose magnitude is now $\alpha$ times the magnitude of $\mathbf{x}$. If $\alpha \lt 0$, then length is scaled the same way and $\alpha\mathbf{x}$ points in the opposite direction of $\mathbf{x}$. Using the physicist picture, this looks like


For the above image, we have vector $\mathbf{x} = \begin{bmatrix}\ \ 3 \\ -2 \end{bmatrix}$,

$$ \frac{1}{2}\mathbf{x} = \frac{1}{2}\begin{bmatrix}\ \ 3 \\ -2 \end{bmatrix} = \begin{bmatrix} \frac{3}{2} \\ -1 \end{bmatrix} $$
and
$$ -\mathbf{x} = (-1)\left[\begin{array}{rr} 3 \\ -2 \end{array}\right] = \left[\begin{array}{rr} -3 \\ 2 \end{array}\right]. $$

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Vector Addition

Vector addition is computed component-wise from the programmer's perspective. If you have two vectors $\mathbf{x}$ and $\mathbf{y}$, their vector sum is

$$\mathbf{x} + \mathbf{y} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} + \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} = \begin{bmatrix} x_1 + y_1 \\ x_2 + y_2 \end{bmatrix}. $$
From the physicist perspective, vector addition is commonly described using the term triangle law. This is because if the vectors $\mathbf{x}$ and $\mathbf{y}$ are arranged so that the tail of $\mathbf{x}$ is at the origin and the tail of $\mathbf{y}$ sits at the tip of $\mathbf{x}$, then the sum of those vectors is the directed line segment from the origin to $\mathbf{y}$'s tip.

The commutative property of vector addition

$$ \mathbf{x} + \mathbf{y} = \mathbf{y} + \mathbf{x} $$
may be demonstrated visually as well by forming a parallelogram

This is simply drawing both possible triangles from the triangle law, switching the ordering of $\mathbf{x}$ and $\mathbf{y}$, with the combination of those two triangles forming a parallelogram. Heuristically, this leads to people describing vector addition as adding the arrows "tip to tail," with the order being irrelevant.

We focused on $\mathbb{R}^2$ here, but the above description and properties work for all $\mathbb{R}^n$, just with more complicated pictures. It isn't truly possible to graph the physicist perspective for four or more dimensions, but the programmer's perspective works just fine for our computations. The mathematician's perspective will become clear soon.

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Linear Combinations

A linear combination is a finite sum of scaled vectors. When one combine's scalar multiplication and vector addition one obtains a linear combination. $$ \mathbf{v} = \alpha_1\mathbf{x} + \alpha_2\mathbf{y} $$

Exercise 2

If $\mathbf{x} = \begin{bmatrix} 1 \\ 3 \end{bmatrix}$ and $\mathbf{y} = \begin{bmatrix} 0 \\ 2 \end{bmatrix}$, what is the vector $2\mathbf{x} - 3\mathbf{y}$?

View Solution

$$ \begin{align*} 2\mathbf{x} - 3\mathbf{y} &= 2\begin{bmatrix} 1 \\ 3 \end{bmatrix} - 3\begin{bmatrix} 0 \\ 2 \end{bmatrix} = \begin{bmatrix} 2 \\ 6 \end{bmatrix} - \begin{bmatrix} 0 \\ 6 \end{bmatrix} = \begin{bmatrix} 2-0 \\ 6-6 \end{bmatrix} = \begin{bmatrix} 2 \\ 0 \end{bmatrix} \\ \end{align*} $$

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4.1.4 Properties of Vector Addition and Scalar Multiplication

In $\mathbb{R}^2$, vector addition and scalar multiplication share many algebraic properties with matrices.

Theorem 4.1 Properties of Vector Addition and Scalar Multiplication on the Plane

Let $\mathbf{u}$, $\mathbf{v}$, and $\mathbf{w}$ be vectors in the plane, and let $c$ and $d$ be scalars.

1. $\mathbf{u} + \mathbf{v}$ is a vector in the plane
2. $\mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u}$
3. $(\mathbf{u} + \mathbf{v}) + \mathbf{w} = \mathbf{u} + (\mathbf{v} + \mathbf{w})$
4. $\mathbf{u} + \mathbf{0} = \mathbf{u}$
5. $\mathbf{u} + (-\mathbf{u}) = \mathbf{0}$
6. $c\mathbf{u}$ is a vector in the plane
7. $c(\mathbf{u} + \mathbf{v}) = c\mathbf{u} + c\mathbf{v}$
8. $(c + d)\mathbf{u} = c\mathbf{u} + d\mathbf{u}$
9. $c(d\mathbf{u}) = (cd)\mathbf{u}$
10. $1(\mathbf{u}) = \mathbf{u}$

Proof (of some of the properties)

Let $\mathbf{u}$, $\mathbf{v}$, and $\mathbf{w}$ be vectors in the plane, and let $c$ and $d$ be scalars.

$$ \begin{align*} 2.\ \ \mathbf{u} + \mathbf{v} &= \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} + \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} \\ &= \begin{bmatrix} u_1 + v_1 \\ u_2 + v_2 \end{bmatrix} \\ &= \begin{bmatrix} v_1 + u_1 \\ v_2 + u_2 \end{bmatrix} \\ &= \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} + \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} \\ &= \mathbf{v} + \mathbf{u} \\ \\ 7.\ \ c(\mathbf{u} + \mathbf{v}) &= c\left(\begin{bmatrix} u_1 \\ u_2 \end{bmatrix} + \begin{bmatrix} v_1 \\ v_2 \end{bmatrix}\right) \\ &= c\begin{bmatrix} u_1 + v_1 \\ u_2 + v_2 \end{bmatrix} \\ &= \begin{bmatrix} c(u_1 + v_1) \\ c(u_2 + v_2) \end{bmatrix} \\ &= \begin{bmatrix} cu_1 + cv_1 \\ cu_2 + cv_2 \end{bmatrix} \\ &= \begin{bmatrix} cu_1 \\ cu_2 \end{bmatrix} + \begin{bmatrix} cv_1 \\ cv_2 \end{bmatrix} \\ &= c\begin{bmatrix} u_1 \\ u_2 \end{bmatrix} + c\begin{bmatrix} v_1 \\ v_2 \end{bmatrix} \\ &= c\mathbf{u} + c\mathbf{v} \end{align*} $$

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4.1.5 Vector Operations in $\mathbb{R}^3$

Let $\mathbf{u} = \langle -4, -2, 1 \rangle$ and $\mathbf{v} = \langle 0, 3, -4 \rangle$

1. Determine the magnitudes of $\mathbf{u}$, $\mathbf{v}$, and $\mathbf{u} + \mathbf{v}$

2. Perform each vector operation

   (a) $2\mathbf{u} + \mathbf{v}$

   (b) $5\mathbf{u}$

   (c) $3\mathbf{u} - 4\mathbf{v}$

View Solution

$$ \begin{align*} 1.\ \ \left\|\langle -4, -2, 1 \rangle\right\| &= \sqrt{ (-4)^2 + (-2)^2 + 1^2 } = \sqrt{16 + 4 + 1} = \sqrt{21} \\ \\ \left\|\langle 0, 3, -4 \rangle\right\| &= \sqrt{ 0^2 + 3^2 + (-4)^2 } = \sqrt{9 + 16} = 5 \\ \\ \left\|\langle -4, -2, 1 \rangle + \langle 0, 3, -4 \rangle\right\| &= \left\|\langle -4, 1, -3 \rangle\right\| \\ &= \sqrt{ (-4)^2 + 1^2 + (-3)^2 } = \sqrt{ 16 + 1 + 9 } = \sqrt{26} \end{align*} $$

$$ \begin{align*} \text{(a)} &\qquad &2\mathbf{u} + \mathbf{v} &= 2\langle -4, -2, 1 \rangle + \langle 0, 3, -4 \rangle = \langle -8, -1, -2 \rangle \\ \\ \text{(b)} &\qquad &5\langle -4, -2, 1 \rangle &= \langle -20, -10, 5 \rangle \\ \\ \text{(c)} &\qquad &3\mathbf{u} - 4\mathbf{v} &= 3\langle -4, -2, 1 \rangle - 4\langle 0, 3, -4 \rangle \\ &\qquad &\ &= \langle -12, -6, 3 \rangle - \langle 0, 12, -16 \rangle \\ &\qquad &\ &= \langle -12, -18, 19 \rangle \end{align*} $$

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4.1.6 Properties of Vector Space $\mathbb{R}^n$

Theorem 4.2 Properties of Vector Addition and Scalar Multiplication in $\mathbb{R}^n$

Let $\mathbf{u}$, $\mathbf{v}$, and $\mathbf{w}$ be vectors in the plane, and let $c$ and $d$ be scalars.

1. $\mathbf{u} + \mathbf{v}$ is a vector in the plane
2. $\mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u}$
3. $(\mathbf{u} + \mathbf{v}) + \mathbf{w} = \mathbf{u} + (\mathbf{v} + \mathbf{w})$
4. $\mathbf{u} + \mathbf{0} = \mathbf{u}$
5. $\mathbf{u} + (-\mathbf{u}) = \mathbf{0}$
6. $c\mathbf{u}$ is a vector in the plane
7. $c(\mathbf{u} + \mathbf{v}) = c\mathbf{u} + c\mathbf{v}$
8. $(c + d)\mathbf{u} = c\mathbf{u} + d\mathbf{u}$
9. $c(d\mathbf{u}) = (cd)\mathbf{u}$
10. $1(\mathbf{u}) = \mathbf{u}$

Proof (of some of the properties)> Let $\mathbf{u}$, $\mathbf{v}$, and $\mathbf{w}$ be vectors in the plane, and let $c$ and $d$ be scalars.

Let $\mathbf{u}$, $\mathbf{v}$, and $\mathbf{w}$ be vectors in the plane, and let $c$ and $d$ be scalars.

$$ \begin{align*} 2.\ \ \mathbf{u} + \mathbf{v} &= \langle u_1, u_2, \dots, u_n \rangle + \langle v_1, v_2, \dots, v_n \rangle \\ &= \langle u_1 + v_1, u_2 + v_2, \dots, u_n + v_n \rangle \\ &= \langle v_1 + u_1, v_2 + u_2, \dots, v_n + u_n \rangle \\ &= \langle v_1, v_2, \dots, v_n \rangle + \langle u_1, u_2, \dots, u_n \rangle \\ &= \mathbf{v} + \mathbf{u} \\ \\ 7.\ \ c(\mathbf{u} + \mathbf{v}) &= c\left(\begin{bmatrix} u_i \end{bmatrix} + \begin{bmatrix} v_i \end{bmatrix} \right) \\ &= c\begin{bmatrix} u_i + v_i \end{bmatrix} \\ &= \begin{bmatrix} c(u_i + v_i) \end{bmatrix} \\ &= \begin{bmatrix} cu_i + cv_i \end{bmatrix} \\ &= \begin{bmatrix} cu_i \end{bmatrix} + \begin{bmatrix} cv_i \end{bmatrix} \\ &= c\begin{bmatrix} u_i \end{bmatrix} + c\begin{bmatrix} v_i \end{bmatrix} \\ &= c\mathbf{u} + c\mathbf{v} \end{align*} $$

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4.1.7 Properties of Additive Identity and Inverse

Theorem 4.3 Properties of Additive Identity and Additive Inverse

Let $\mathbf{v}$ be vectors in the plane, and let $c$ be scalars.

1. If $\mathbf{v} + \mathbf{u} = \mathbf{v}$, then $\mathbf{u} = \mathbf{0}$
2. If $\mathbf{v} + \mathbf{u} = \mathbf{0}$, then $\mathbf{u} = -\mathbf{v}$
3. $0\mathbf{v} = \mathbf{0}$
4. $c\mathbf{0} = \mathbf{0}$
5. If $c\mathbf{v} = \mathbf{0}$, then $c=0$ or $\mathbf{v}=\mathbf{0}$
6. $-(-\mathbf{v})= \mathbf{v}$

Proof (of some of the properties)

Let $\mathbf{u}$, and $\mathbf{v}$ be vectors $\mathbf{R}^n$, and let $c$ a scalar.

2. If $\mathbf{v} + \mathbf{u} = \mathbf{0}$, then using Theorem 4.2 we have
   $$ \begin{align*} \mathbf{v} + \mathbf{u} &= \mathbf{0} &\qquad\text{Given} \\ -\mathbf{v} + \left( \mathbf{v} + \mathbf{u} \right) &= -\mathbf{v} + \mathbf{0} &\qquad\text{Theorem 4.2.1} \\ -\mathbf{v} + \left( \mathbf{v} + \mathbf{u} \right) &= -\mathbf{v} &\qquad\text{Theorem 4.2.4} \\ \left( -\mathbf{v} + \mathbf{v} \right) + \mathbf{u} &= -\mathbf{v} &\qquad\text{Theorem 4.2.3} \\ \mathbf{0} + \mathbf{u} &= -\mathbf{v} &\qquad\text{Theorem 4.2.5} \\ \mathbf{u} + \mathbf{0} &= -\mathbf{v} &\qquad\text{Theorem 4.2.2} \\ \mathbf{u} &= -\mathbf{v} &\qquad\text{Theorem 4.2.4} \\ \end{align*} $$
   

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