Vector spaces

A vector, sometimes called a linear, space is a set of elements that must satisfy certain properties that make the elements of this set, which we call vectors, behave "nicely".

An example of a vector space

Before we define a vector space, let's show an example of a vector space that you should already know from high school, only you don't know that it's a vector space. These are classical vectors in the plane.

By an in-plane vector, we mean an ordered pair [a, b], where a, b ∈ ℝ. are any coordinates in the space ℝ² (meaning the Cartesian product ℝ×ℝ). We can represent such a vector by an arrow from a point [0, 0] to a point [a, b]. The vector [2, 3] would look like this:

We know that we can add vectors according to the rule [a, b] + [c, d] = [a + c, b + d]. The sum of the vectors [2, 3] + [4, 1] would be equal to the vector [2 + 4, 3 + 1] = [6, 4]. We would represent it graphically as follows:

At the same time, we can compute the k-fold of the vector, where k is a real number. k-fold is written as k · [a, b]. The result is a new vector, which is defined as k · [a, b] = [k · a, k · b]. Graphically, this is shown by the vector being "stretched", "shortened", or changing direction. If we have a vector u = [2, 3], then the vectors 2 · u, $\frac12u$ and −1 · u look like this:

We will prove later that the set of all vectors ℝ² with operations defined in this way form a vector space. First, we will define a vector space and use this vector space ℝ² to show each property.

Definition of vector space

A vector space (or linear space) is a nonempty set V, the elements of which are called vectors. Furthermore, two operations must exist on the set: vector addition, i.e., the $V \times V \longrightarrow V$ representation, and vector multiplication by a real number, i.e., the $\mathbb{R} \times V \longrightarrow V$ representation. (More generally, any solid can be used in place of a set of real numbers, but more on that in later articles.) We denote vector addition by +, vector multiplication by · . These two operations must satisfy the following conditions. That is, they must hold for all x, y, z ∈ V and for all a, b ∈ ℝ (note that the vectors x, y, z are in bold, while the numbers a, b are normal):

1. x + y = y + x: vector addition is commutative. Returning to the example with the vectors in ℝ², it doesn't matter if we add the vectors like this [2, 3] + [4, 1] or vice versa: [4, 1] + [2, 3] .

• (x+y)+z = x+(y+z): vector addition is associative. Again, we can plug in some vectors from ℝ² and see that it works.
• a · (b · x) = (a · b) · x: associative multiplication law. For our vectors, it will be true that 4 · (5 · [6, 7]) is the same as (4 · 5) · [6, 7]. Both examples lead to the vector [20 · 6, 20 · 7].
• a · (x + y) = a · x + a · yThe : distributive law for vectors from V. Again we see that 2 · ([2,3]+[4,5]) is the same as 2 · [2,3]+2 · [4,5]. Both expressions lead to the vector [2 · 6,2 · 8].
• (a + b) · x = a · x+b · x: distributivity of addition of numbers.
• 1 · x = x: if we multiply any vector by the real number 1, the value of the vector does not change. In our case, 1 · [5, 8] = [5,8].
• There exists an element 0∈ V such that 0 · x = 0. A bold zero, i.e. 0, is a so-called zero vector, while a non-bold zero 0 is simply the number zero. This point says that there must be a vector in V, which we obtain by multiplying any vector x∈ V by the number zero. For our space ℝ², this is the vector [0,0], because for all vectors [a, b] ∈ ℝ², it is true that 0 · [a,b] = [0,0].

These are all properties that a vector (linear) space V must satisfy. If we multiply a vector by some real number a, then we call that number a a scalar. So we often speak of multiplying by a scalar and mean an expression of the form a · x, where a is a real number and x is a vector.

The double meaning of the plus sign +

Note that in some expressions the plus sign + can have a double meaning. For example, in the distributivity point of addition of numbers, we have the equation (a + b) · x = a · x+b · x. Yet the + sign on the left side of the equation adds real numbers because a and b are both real numbers, while on the right side it adds vectors because the product a · x is a vector.

This automatic recognition is often used in mathematics - it is always clear from the context what addition is meant. If we have the expression x + y and yet x, y are vectors, we use vector sum. If x, y were numbers, we would use a classical sum. If they were matrices, we would use matrix sum, etc.

We could use a different sign for vector addition, for example $\oplus$. Then we could write the distributivity of number addition as follows: $(a+b)\cdot \mathbf{x} = a\cdot \mathbf{x}\oplus b\cdot \mathbf{x}$. The normal + here denotes addition of real numbers a, b, the sign $\oplus$ denotes addition of vectors. But usually a single + sign is used for both versions of addition, and its meaning is known from the context. The same applies to the · sign.

Vector space R²

In the first section we defined the vector space V = ℝ², now we will prove that it is indeed a vector space. The first thing we need to check is that we have defined the addition and multiplication operations correctly. Addition is supposed to add two vectors and the result is supposed to be a new vector. This is satisfied because [a, b]+[c, d] = [a + c,b + d]. Adding the two vectors from ℝ² results in a new vector from ℝ².

If, for example, we defined the addition of vectors as the sum of their magnitudes, then this would not be a valid sum operation in vector space. We would compute the size of the vector [a, b] as $\sqrt{a^2+b^2}$. Thus the result of adding [3,4]+[5, 12] would be the number $\sqrt{3^2+4^2}+\sqrt{5^2+12^2}=5+13=18$, which is not a vector of the space ℝ².

Next, we need to verify that we have defined the multiplication correctly. Multiplication multiplies a real number with a vector from ℝ² and the result should be a new vector from ℝ². This is again satisfied because we have defined scalar multiplication as k · [a,b] = [k · a, k · b].

We will check all 7 properties that a vector space must satisfy one by one. We will substitute our vectors [a, b], [c, d], … for the individual vectors x, y as appropriate, don't be afraid of that.

1. x + y = y + x, commutativity of vector addition. Is our defined vector addition commutative? Yes, it is, because it is defined as [a, b]+[c, d] = [a + c,b + d]. If we swap the vectors on the left, we get: [c, d]+[a, b] = [c + a,d + b]. Since c + a is the sum of two real numbers and the sum of real numbers is a commutative operation, c + a = a + c and the same for the other pair: d + b = b + d. Thus, it holds that [a + c,b + d] = [c + a,d + b].

• (x+y)+z = x+(y+z). This is going to be kind of playing with parentheses. Basically, these are simple modifications, only they will be messy because there will be too many parentheses. Let's try to calculate this vector sum:

$$ \left(\left[a,b\right]+\left[c,d\right]\right)+\left[e,f\right] = \left[a+c,b+d\right]+\left[e,f\right]=\left[\left(a+c\right)+e, \left(b+d\right)+f\right] $$

Since a, c, e are real numbers, and the addition operation is associative on real numbers, (a + c)+e = a + c + e. Same for the second coordinate. So we can write that

$$ \left(\left[a,b\right]+\left[c,d\right]\right)+\left[e,f\right] = \left[a+c+e, b+d+f\right] $$

Now let's try to add the same vectors with differently stacked parentheses:

$$\left[a,b\right]+\left(\left[c, d\right]+\left[e, f\right]\right)=\left[a,b\right]+\left[c+e,d+f\right]=\left[a+\left(c+e\right),b+\left(d+f\right)\right]$$

Again, a+(c + e) = a + c + e, so we can write that

$$\left[a,b\right]+\left(\left[c, d\right]+\left[e, f\right]\right)=\left[a+c+e, b+d+f\right]$$

Thus we see that no matter how we bracket the sum of the three vectors, we always arrive at the same vector [a + c + e, b + d + f]. The addition operation in ℝ² is thus associative and satisfies the second property of vector spaces.

You can try substituting specific numbers after the a, …, f variables and you'll see that it works.

• a · (b · x) = (a · b) · x. Again, we first break down the left side of the equation, substituting the vector [c, d] after x:

$$a\cdot \left(b \cdot \left[c, d\right]\right) = a \cdot \left[b \cdot c, b \cdot d\right] = \left[a \cdot (b \cdot c), a\cdot (b\cdot d)\right]$$

Since multiplication of real numbers is associative, we get the equation a · (b · c) = a · b · c. Thus we can write:

$$a\cdot \left(b \cdot \left[c, d\right]\right) = \left[a\cdot b \cdot c,a\cdot b\cdot d\right]$$

We now decompose the right-hand side of the equation, i.e., the side (a · b) · x:

$$(a\cdot b)\cdot\left[c, d\right]=\left[(a\cdot b)\cdot c, (a\cdot b)\cdot d\right]$$

Again, (a · b) · c = a · b · c, so:

$$(a\cdot b)\cdot\left[c, d\right]=\left[a\cdot b \cdot c,a\cdot b\cdot d\right]$$

We see that both sides of the equation lead to the resulting vector [a · b · c,a · b · d], so the equation holds.

• a · (x + y) = a · x + a · y. The procedure will be the same, let's break down the left side of the equation:

$$a\cdot\left(\left[c, d\right]+\left[e, f\right]\right) = a\cdot\left(\left[c+e, d+f\right]\right)=\left[a\cdot(c+e), a\cdot (d+f)\right]$$

Since a, c, e are real numbers where the distributive law holds, we get the equation: a · (c + e) = ac + ae. Thus we can write:

$$a\cdot\left(\left[c, d\right]+\left[e, f\right]\right) = \left[ac+ae, ad+af\right]$$

The right side of the equation:

$$a\cdot\left[c,d\right]+a\cdot\left[e,f\right]=\left[ac, ad\right]+\left[ae, af\right]=\left[ac+ae, ad+af\right]$$

We can see that the left and right sides of the equation are equal after adjustments.

• (a + b) · x = a · x+b · x: We will proceed very similarly to the previous step. Adjusting the left side of the equation:

$$(a+b)\cdot\left[c, d\right]=\left[(a+b)\cdot c,(a+b)\cdot d\right]=\left[ac+bc,ad+bd\right]$$

Right side:

$$a\cdot\left[c, d\right]+b\cdot\left[c,d\right]=\left[ac, ad\right]+\left[bc, bd\right]=\left[ac+bc,ad+bd\right]$$

• 1 · x = x: I don't believe anyone has even made it to this point, but so be it, we'll move on :-). At least this point is easy. We have to verify that 1 · [a, b] = [a,b]. This is certainly true, because

$$1\cdot\left[a,b\right]=\left[1\cdot a, 1\cdot b\right]=\left[a,b\right]$$

• There is an element 0∈ ℝ² such that 0 · x = 0: is a vector [0,0]. For it, it holds that 0 · [a,b] = [0,0], because

$$0\cdot\left[a,b\right]=\left[0\cdot a, 0\cdot b\right]=\left[0{,}0\right]$$

The size of vector spaces

What is the smallest vector space that exists? By definition, we know that a vector space is a nonempty set V, so the smallest possible size we can think of is a set of one vector. From the seventh point of the definition, we know that the space must contain a zero vector. The set V = {0} is thus a candidate for the smallest vector space.

We can define addition as 0+0 = 0 and multiplication as a · 0 = 0. No matter what we do, we always get a zero vector. Obviously, this one-point set will have all seven required properties.

We already know that there is a one-point vector space. Is there a vector space that will have two elements? Can we define addition and multiplication operations on the set V = {x, 0} to produce a vector space?

The answer is no, there is no two-point vector space. Nor is there any finite vector space other than the aforementioned one-point vector space. All other vector spaces are already infinite. But if, as has been suggested, we use some other, finite, solid instead of the set of real numbers, we could get another finite vector space.

Summary

The definition of vector spaces is not entirely straightforward, vectors and operations must satisfy a total of 7 conditions, which is not a few. On the other hand, these conditions are quite trivial and understandable - they aim at making vectors easy to work with. Thanks to the defined conditions, we can add vectors from a vector space without having to check if the result of the addition will be a vector from another vector space. Similarly, we can multiply a vector by a real number and always get a vector from the same space. That vectors and their operations obey commutative, associative and distributive laws is also quite natural.

We could also define vector space more generally. Instead of a set of real numbers, we can take any solid. But more on that in later articles.

Links and resources

• O. Linear Algebra 1 [PDF]
• Petr Olšák: Linear Algebra