I can't understand this paragraph on vector spaces. Can anyone help demistify this?

Question

A collection of vectors $u_1, \cdots, u_n$ in $V$, spans $V$ if every element in $V$ can be written as a linear combination of the $u$’s; that is, $$v = a_1u_1 + \cdots + a_nu_n \quad \quad (2.10)$$

where $a_1, \cdots, a_n$ are complex numbers. The vectors $u_1 \cdots u_n$ are called linearly independent if $v = a_1u_1 + \cdots + a_nu_n = 0$ implies $a_1 = a_2 = \cdots = a_n = 0$. If $u_1, \cdots, u_n$ are linearly independent and span $V$ they are called a basis of $V$. The number $n$ is unique and is called the dimension of $V$. Suppose that $W$ is a collection of vectors from a vector space $V$. $W$ is a subspace of $V$ if: (1) for every $v, w$ in $W$, $v + w$ is also in $W$; (2) for every $w$ in $W$ and every scalar $\alpha$, $\alpha v$ is also in $W$.

So far from what I understand, a vector space is essentially an abstraction of the idea/properties of a vector that can be applied freely to many things.

This paragraph is right after talking about the vector space over $\mathbb{C}$ and I can’t wrap my head around what its trying to get across.

Answer 1

The idea is that you can build a vector space from a fixed number of vectors.

The number plane is a vector space. Define an $(x,y)$ coordinate system on it. You can build the whole number plane from the two vectors $(1,0)$ and $(0,1)$. It is easy to show that any vector $(a,b)$ can be written as $a(1,0) + b(0,1)$. The vectors $(1,0)$ and $(0,1)$ span the number plane. Another way to say it is $(1,0)$ and $(0,1)$ are a basis for the number plane.

It turns out there are many ways to choose a basis for the number plane. You can pick $(1,0)$ and $(1,1)$ as your basis, or perhaps $(2,2)$ and $(3,4)$.

You cannot choose $(1,0)$ and $(2,0)$. One of these is a multiple of the other. They are not linearly independent. These vectors can only build the $x$ axis. Given any vector on the $x$ axis, there are many ways to build it from these to vectors.

So these things are all the same

• The entire number plane can be built from two linearly independent vectors in the number plane.
• Two linearly independent vectors in the number plane span the number plane.
• The dimension of the number plane is two.

You can do similar things from the $3$-D vector space $(x,y,z)$. The dimension of this space is $3$ because it takes $3$ vectors to span it.

Answer 2

Indeed the concept of vector space is an abstraction that can apply to many things, beyond even what is set forth in the book.

I would assume that if you are in a physics course that is going to seriously study any aspect of quantum mechanics (at least as far as needing the kind of vector space introduced in that passage in the book), you have already completed at least one physics course that invoked displacement vectors, velocity vectors, and acceleration vectors, both in the $x,y$ Cartesian plane (also known as $\mathbb R^2$) and in $x,y,z$ Cartesian space (also known as $\mathbb R^3$).

The vectors that you used in the Cartesian plane and Cartesian space were just as much members of vector spaces as anything described in the quoted passage. You should already have formed linear combinations of vectors within these vector spaces, and if you have ever made particular use of the vectors with coordinates $(1,0,0),$ $(0,1,0),$ and $(0,0,1),$ perhaps naming them $\mathbf i,$ $\mathbf j,$ and $\mathbf k$, you have used a basis of the vector space $\mathbb R^3.$

One thing that is different about the vector spaces $\mathbb R^2$ and $\mathbb R^3$ from the vector spaces in the book are that the coordinates of vectors in $\mathbb R^2$ and $\mathbb R^3$ are only real numbers, and likewise you can only use real numbers as scalars to multiply those vectors by. In the book, you have vectors whose coordinates are complex numbers and you can use complex scalars. Technically, the definition given there is not a general definition of vector space, but rather is a definition of complex vector space. Also, the book seems to be setting up to use more than three coordinates in a vector. Much of the mechanics of manipulating the vectors will be similar, however.

One difference between physicists and mathematicians is that physicists tend not to be as interested in the rigorous underpinnings of their mathematics as long as the mathematics apparently works for whatever purpose they need it for. So it would not be surprising to find a definition of vector space in a physics textbook (assuming this is one) that elicits the kind of complaints you see in the comments. If you want to look into this further, you could look up "vector spaces" and then "complex vector spaces", trying to stick to mathematical sources.

Or you might try forging ahead to see if the author of this book clarifies things later.

Answer 3

There is a branch of mathematics known as linear algebra.

That paragraph you quoted is amounts to condensing a $500$ page long textbook on linear algebra condensed into a single paragraph.

• People familiar with linear algebra do not need to be told what the basis and "span of a basis" are, they already know.

• People not familiar with linear algebra will have no idea what that guy or gal is talking about.

As such, the paragraph serves no useful purpose.

I will try to explain what some of that stuff is:

• linear combination
• linearly independent
• span
• basis
• dimension
• vector space
• subspace

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Walking in the desert

Imagine standing in a very flat dessert. Instead of being allowed to walk wherever you wish, you are told that you may only walk in straight lines in certain directions. However, after walking for awhile, you are allowed to stop, and choose a different direction from the set of directions which are allowed, and walk in that direction. In the end you kind of "zig-zag" your way across the dessert.

A basis is a set of directions which are basically, the minimum number of directions you need to theoretically go anywhere in the desert.

Suppose that your present position in a very flat desert is record as some x-y coordinates.

If the only thing you could do was add $(3.2, 4.1)$ to your present position over and over again, then you probably would be able to get everywhere in the dessert that you want to.

Suppose that, instead, you had more choices. The allowed directions are $B = \left \{\dfrac{(+0.2, -4.9)}{4.90408}, \dfrac{(+1.1, +\pi)}{\vert (+1.1, \pi) \vert} \right \}$.

The notation $\vert (+1.1, \pi) \vert$ is equivalent to:

• the length of the vector $(+1.1, \pi)$
• the length of the hypotenuse of a right-triangle, where the legs of the triangle are $+1.1$ and $+\pi$

Now, pick a random direction from the set of allowed directions. Maybe you choose $\dfrac{(+1.1, +\pi)}{\vert (+1.1, \pi) \vert}$.

Next, pick a random distance $d$. Counterintuitively, the distance $d$ is allowed to be a negative number, such as $-4.8$.

If your current location is $(x, y)$, then your next stop will be $(x, y) + d* \dfrac{(+1.1, +\pi)}{\vert (+1.1, \pi) \vert}$

The span of a basis is the set of all locations in the dessert you could potentially get to by taking turns in the directions that are allowed.

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Chess

Consider a chess board.

You do not need to know all of the rules to chess in order to understand the following discussion.

There a few pieces on the board, named "bishops", which are only allowed to move diagonally. That is, a Bishop on a black square must always stay on black squares.

Imagine that a bishop is the only piece on the board.

The bishop was initially on a black square.

For a bishop, maybe some allowable directions are $(+1, +1)$ and $(-1, +1)$

An example of a linear combination of $(-1, +1)$ and $(-1, +1)$ is $4*(-1, +1)+7*(-1, +1)$

A linear combination of some directions is basically a new complicated direction formed by combining the basic/simple directions you started with.

Then the set of all of the black squares on the board is an example of one vector space

The entire chess board is another example of a vector space.

A basis of the black squares is a minimal set of direction vectors, such that those few directions are still enough to get to any black square on the board.

A basis of the entire chess board is the simplest set of direction vectors a mathematician can think of, such that that those few short, simple directions are enough to get to anywhere on the chess board.

The black squares are a subspace of the entire board.

The span of set of directions $\mathtt{DIRS}$ is the set of all squares on the chess board you can get to by using only the directions listed/specified in $\mathtt{DIRS}$.

A chess board can be modeled as the set of all coordinate pairs $(r, c)$ such that $r, c \in \mathbb{N} \text{ and } 1\leq r,c \leq 8$

$r, c \in \mathbb{N}$ is equivalent to:
"$r$ and $c$ are whole numbers, such as $1$, $2$, $3$, etc..." but $r$ and $c$ are NOT decimal numbers such as $\pi \approx 3.14 \cdots$

For example, $(5, 7)$ is a square on the board.

Suppose that $\mathtt{BOARD} = \left \{(r, c) \text{ : } r, c \in \mathbb{N} \text{ and } 1\leq r,c \leq 8 \right \}$

I was going somewhere with that, but I'm tired, and do not want to write anymore. Hopefully that helps a little bit.

Answer 4

The text you give is poorly typeset (in the image in your original version of your question) and a little carelessly written, which makes it harder to understand. Perhaps it is from some lecture notes or the notes of a student from a lecture course.

What the text says

The text defines six fundamental concepts in vector spaces:

• The first sentence defines the terms linear combination and span (though that is a little unclear).
• The second sentence defines the term linearly independent.
• The third sentence defines the term basis using the terms span and linearly independent.
• The fourth sentence defines the term dimension as the number of elements in any basis.
• The fifth and sixth sentences define the term subspace.

These tell you almost nothing new about vector spaces, but they are very important for understanding what will follow.

You say “This paragraph is right after talking about the vector space over $ \mathbb{C} $”, by which I expect you mean just before that paragraph there is a definition of vector space and perhaps some examples.

Why we study vector spaces

You are right to think that a vector space is an abstraction from the properties of the sort of vectors that we use to describe positions on a plane (in 2-D) or in 3-D space. There are indeed many other things that have enough in common with such vectors to make it worth studying them in the abstract.

Using the same terminology for all vector space also makes them easier to understand.

Clarification of the text

If you realise what is being defined, you should be able to understand the definitions with some effort — but do not rush! Mathematical texts that define a lot of terms take some time to absorb, and as long as you are stuck on one definition you may not be able to make sense of the next.

I have added extra definitions (scalar, coefficient and depends on) to make the conceptual steps smaller and more obvious.

Extra definition: “scalar”

By a scalar we mean anything from the “field” over which the vector space is defined; in your case a scalar is a complex number (from $ \mathbb {C} $), and in physics you may also find yourself dealing with vector spaces over $ \mathbb {R} $, the real numbers. The definitions that follow apply to all vector spaces, even those over quite different scalar fields.

First sentence

Original:

A collection of vectors $u_1, \cdots, u_n$ in $V$, spans $V$ if every element in $V$ can be written as a linear combination of the $u$’s; that is, $$v = a_1u_1 + \cdots + a_nu_n \quad \quad (2.10)$$

where $a_1, \cdots, a_n$ are complex numbers.

A linear combination of vectors $u_i$ means a sum of scalar multiples of those vectors, i.e. anything which can be written in the form $\alpha_1u_1 + \cdots + \alpha_nu_n $, where the $\alpha_i$ are scalars. We also call the $ \alpha_i $ coefficients.

A vector $ v $ depends on a collection $ U $ if $v$ can be written as a linear combination of the elements $ u_i $ of $ U $.

A collection of vectors $ U $ is said to span a vector space $ V $ if every vector in $ V $ can be written as a linear combination of the vectors in $ U $, i.e. if every vector in the space $ V $ depends on the collection $ U $.

Second sentence

Original:

The vectors $u_1 \cdots u_n$ are called linearly independent if $v = a_1u_1 + \cdots + a_nu_n = 0$ implies $a_1 = a_2 = \cdots = a_n = 0$.

A collection of vectors $ U $ is said to be linearly independent if no one of them depends on the rest of the collection.

This is equivalent to saying that the only linear combination of $ U $ that yields the zero vector is the combination with all coefficients equal to the scalar 0. This formulation is usually preferred, perhaps because it does not single out one element of the collection.

To see that the definitions are equivalent, if $ u_1 = \alpha_2 u_2 + \alpha_3 u_3 $, we can conclude $ -1 u_1 + \alpha_2 u_2 + \alpha_3 u_3 = 0 $ .

Conversely, if $ \alpha_1 u_1 + \alpha_2 u_2 + \alpha_3 u_3 = 0 $, with $ \alpha_1 \neq 0 $, we can conclude $ u_1 = \frac {- \alpha_2} {\alpha_1} u_2 + \frac {- \alpha_3} {\alpha_1} u_3 $.

Third sentence

Original:

If $u_1, \cdots, u_n$ are linearly independent and span $V$ they are called a basis of $V$.

A basis of a vector space means a minimal (smallest possible) collection that spans the space.

If a $ U $ spans $ V $ but is not linearly independent, we can span $ V $ with a smaller collection by removing anything that depends on other elements and rewriting it in terms of those it depends on. Hence a basis must be linearly independent, as otherwise it would not be minimal.

It turns out that all bases of a given vector space have the same number of elements. This is what the fourth sentence means by “The number $n$ is unique”. This can be proven, but you may be expected to take it on trust.

Fourth sentence

Original:

The number $n$ is unique and is called the dimension of $V$.

The dimension of a vector space means the number of elements in any basis of that space.

We have just observed that this is well defined.

I think that, at least to start with, you will only be concerned with finite-dimensional vector spaces, i.e. those with a finite basis. Infinite-dimensional vector spaces also exist.

Fifth and sixth sentence

Original:

Suppose that $W$ is a collection of vectors from a vector space $V$. $W$ is a subspace of $V$ if: (1) for every $v, w$ in $W$, $v + w$ is also in $W$; (2) for every $w$ in $W$ and every scalar $\alpha$, $\alpha v$ is also in $W$.

A subspace of a vector space means a collection of vectors from that space that also forms a vector space using the same vector addition and the same scalar multiplication (by the same scalars).

Since in any vector space, vector addition and scalar multiplication have to yield elements of the space, a subspace has to be “closed” under these operations. This is what the conditions (1) and (2) say.