Part 1 · Vectors

1.1 Points, vectors, bases

Source: Favata’s notes, appendix A.1, p. 10 (points and vectors); the rest is background the notes assume

Objectives

Tell points from vectors, and say why only vectors can be added.
Understand linear combinations, span and linear independence as pictures first.
Recipe: write a vector as a combination of two others, or show that it is not one.
Pictures: the span of one vector, of two independent vectors, of two dependent ones.
Vocabulary: translation space, position vector, linear combination, span, linearly (in)dependent, basis, dimension.

Continuum mechanics describes a body by where its points are, and how far and in which direction they move. Those are two kinds of object, points and vectors, and the book starts by keeping them apart. Most of this section is the linear algebra the lecture notes take as known. If it is familiar, skim the boxes and the demo, then go on to §1.2 Scalar product and components.

Points and vectors

Definition 1.1.1Point space and translation space

We work in a Euclidean point space $E^{n}$ , $n = 1, 2, 3$ ; when $n$ is not stated, $n = 3$ . Its elements are points $P, Q, \dots$

The difference of two points is a vector, an element of the translation space $V$ of $E$ : $P - Q = v \in V .$

The vector $P - Q$ is the arrow that carries $Q$ to $P$ . Vectors can be added and scaled: $V$ is a vector space. Points cannot. “ $P + Q$ ” has no meaning, because it would change with the choice of origin. What makes sense is a point plus a vector, $Q + v = P$ , which moves $Q$ along the arrow.

Definition 1.1.2Position vector

With an origin $O$ fixed, the position vector of a point $P$ is $P - O = OP = p \in V .$ “The point $p$ ” is shorthand for the point at position $p$ from $O$ .

Warning

A position vector depends on the origin; a difference of positions does not: $(P - O) - (Q - O) = P - Q$ . This is why the deformation gradient, which acts on differences of positions (small material segments), will not care where the origin is, while the deformation itself will.

Notation: vectors are bold lowercase ( $u, v$ , sometimes Greek, $ω$ ), scalars are italic, spaces are script ( $E$ , $V$ ). Definitions use $:=$ .

Linear combinations and span

Definition 1.1.3Linear combination, spanbeyond the notes

A linear combination of $v_{1}, \dots, v_{k}$ is any vector $α_{1} v_{1} + α_{2} v_{2} + \dots + α_{k} v_{k}, α_{i} \in R .$ The set of all of them is the span, $span {v_{1}, \dots, v_{k}}$ .

Pictures first:

$span {v}$ for $v \neq = 0$ is the line through the origin along $v$ : all its multiples.
$span {u, v}$ is a plane through the origin, if $u$ and $v$ point in different directions. If they are parallel, it is only a line.
$span {0} = {0}$ , a point.

Recipe 1.1.4Is w a combination of u and v?beyond the notes

In the plane, with $u = (u_{1}, u_{2})$ , $v = (v_{1}, v_{2})$ , $w = (w_{1}, w_{2})$ :

Write $w = α u + β v$ component by component: two equations, $α u_{1} + β v_{1} = w_{1}$ and $α u_{2} + β v_{2} = w_{2}$ .
If $D := u_{1} v_{2} - u_{2} v_{1} \neq = 0$ , there is exactly one solution: $α = (w_{1} v_{2} - w_{2} v_{1}) / D$ , $β = (u_{1} w_{2} - u_{2} w_{1}) / D$ .
If $D = 0$ , $u$ and $v$ are parallel. Then $w$ is a combination only if it lies on their line, and then in infinitely many ways.

The number $D$ is the signed area of the parallelogram on $u, v$ . It comes back as the vector product (§1.3 Vector product and volume) and as the determinant (§2.5 Inverse and determinant).

Loading the demo…

Figure 1.1.1 Drag

u

v

and

w

. The readout writes

w

as a combination, and the dashed path walks it: so many

u

, then so many

v

. Now drag

v

onto the line of

u

(it snaps): the shaded plane collapses to a line, and

w

can no longer be reached unless it is on it.

Linear independence

Definition 1.1.5Linear independencebeyond the notes

Vectors $v_{1}, \dots, v_{k}$ are linearly independent if the only combination that gives $0$ is the trivial one: $α_{1} v_{1} + \dots + α_{k} v_{k} = 0 ⟹ α_{1} = \dots = α_{k} = 0.$ Otherwise they are linearly dependent.

Key observation 1.1.6beyond the notes

A set is dependent exactly when one of its vectors is a combination of the others: if $α_{j} \neq = 0$ in a nontrivial combination equal to $0$ , divide by $α_{j}$ and solve for $v_{j}$ . Removing that vector does not shrink the span.

Two vectors are dependent iff they are parallel (or one is $0$ ).
Three vectors in space are dependent iff they lie in one plane.

Warning

“This vector is linearly dependent” means nothing: independence is a property of a set of vectors. And a dependent set need not have every vector in the span of the others, only at least one: in ${e_{1}, 2 e_{1}, e_{2}}$ , $e_{2}$ is not a combination of the other two.

Basis and dimension

Definition 1.1.7Basis, dimension, componentsbeyond the notes

A basis of $V$ is a linearly independent set that spans $V$ . All bases of $V$ have the same number of elements, the dimension $n$ of $V$ . Every $v$ is then a combination of the basis vectors $b_{1}, \dots, b_{n}$ in exactly one way, $v = v_{1} b_{1} + \dots + v_{n} b_{n},$ and the numbers $v_{i}$ are the components of $v$ in that basis.

Why only one way. If $v = \sum v_{i} b_{i} = \sum v_{i}^{'} b_{i}$ , then $\sum (v_{i} - v_{i}^{'}) b_{i} = 0$ , and independence forces $v_{i} = v_{i}^{'}$ .

In the plane any two non-parallel vectors are a basis; in space, any three that do not lie in one plane. The demo above shows the plane case: while $u, v$ are independent, the readout always finds exactly one pair $α, β$ .

beyond the notes

Components depend on the basis; the vector does not. Most of this book is about quantities that do not depend on the basis either: lengths, angles, the trace, the determinant, eigenvalues. Those are the ones with a physical meaning.

The next section adds a scalar product to $V$ . It makes the most convenient bases available, the orthonormal ones, in which a component is a single scalar product.

Exercises

Exercise 1.1.1beyond the notes

Show that $u = (1, 2)$ , $v = (3, 1)$ are a basis of the plane, and find the components of $w = (5, 5)$ in it.

Solution

$D = 1 \cdot 1 - 2 \cdot 3 = - 5 \neq = 0$ , so they are independent, and two independent vectors in a 2D space are a basis. $α = (5 \cdot 1 - 5 \cdot 3) / (- 5) = 2$ , $β = (1 \cdot 5 - 2 \cdot 5) / (- 5) = 1$ . Check: $2 (1, 2) + (3, 1) = (5, 5)$ .

Exercise 1.1.2beyond the notes

Are $(1, 0, 1)$ , $(0, 1, 1)$ , $(1, 1, 2)$ linearly independent? What is their span?

Solution

The third is the sum of the first two, so they are dependent. The first two are not parallel, so the span is the plane through the origin containing them, the plane $x_{3} = x_{1} + x_{2}$ .