0.1 Fundamental Properties
Let’s first start by solving some problems.
Exercise 0.1.1
Identify if the set \(\mathbb {S} = \{ (x, y) \in \mathbb {R}^2 \mid x + y = 5 \}\) is a vector space.
Solution.
We could see if the set satisfies all axioms for a vector space. However, just by looking at the condition \(x + y = 5\), we know that there cannot exist a zero vector since \(0 + 0 \neq 5\). Because the set violates A4, the set is not a vector space.
Exercise 0.1.2
Identify if the set \(\mathbb {S} = \{ \langle x, y \rangle \in \mathbb {R}^2 \mid x + y = 0 \}\) is a vector space.
Solution.
Unlike the previous problem, we know that there exists a zero vector. For this one, let’s go through each axiom. Let \(u = \langle x_1, -x_1 \rangle \) and \(v = \langle x_2, -x_2 \rangle \). Notice that \(u \oplus v = \langle x_1 + x_2, -x_1 - x_2 \rangle \in \mathbb {R}^2\) and satisfies the condition \(x_1 + x_2 + (-x_1 -x_2) = 0\). Therefore, A1 is satisfied. Moreover, consider the following equations for A2 and A3 where \(w = \langle x_3, -x_3 \rangle \). \begin{align*} u \oplus v &= \langle x_1, -x_1 \rangle + \langle x_2, -x_2 \rangle = \langle x_1 + x_2, -x_1 - x_2 \rangle \\ &= \langle x_2 + x_1, -x_2 - x_1 \rangle = \langle x_2, -x_2 \rangle + \langle x_1, -x_1 \rangle \\ &= v \oplus u \\ u \oplus (v \oplus w) &= \langle x_1, -x_1 \rangle + \left ( \langle x_2, -x_2 \rangle + \langle x_3, -x_3 \rangle \right ) \\ &= \langle x_1, -x_1 \rangle + \langle x_2 + x_3, -x_2 - x_3 \rangle \\ &= \langle x_1 + x_2 + x_3, -x_1 - x_2 - x_3 \rangle \\ &= \langle x_1 + x_2, -x_1 - x_2 \rangle + \langle x_3, -x_3 \rangle \\ &= (u \oplus v) \oplus w \end{align*}
Therefore, A2 and A3 are satisfied. Furthermore, notice that \(\langle 0, 0 \rangle \in \mathbb {R}^2\) and \(0 + 0 = 0\), satisfying A4. Lastly for vector addition A5 holds as \(\langle -x, x \rangle \) is the additive inverse of \(\langle x, -x \rangle \) that satisfies the provided conditions. Thus, all axioms for vector addition are met.
Continuing with scalar multiplication, notice that S1 and S3 satisfy as \(\langle x, y \rangle \in \mathbb {R}^2\) and \(1 \odot \langle x, y \rangle = \langle x, y \rangle \). Moreover, consider the following equations for \(\alpha , \beta \in \mathbb {R}\). \begin{align*} \alpha \odot (\beta \odot u) &= \alpha \odot \langle \beta x_1, -\beta x_1 \rangle = \langle \alpha \beta x_1, -\alpha \beta x_1 \rangle = \alpha \beta \odot \langle x_1, -x_1 \rangle \\ &= (\alpha \beta ) \odot u \\ (\alpha + \beta ) \odot u &= \langle (\alpha + \beta ) x_1, -(\alpha + \beta ) x_1 \rangle = \langle \alpha x_1 + \beta x_1, -\alpha x_1 - \beta x_1 \rangle \\ &= \langle \alpha x_1, -\alpha x_1 \rangle + \langle \beta x_1, -\beta x_1 \rangle \\ &= \alpha \langle x_1, -x_1 \rangle + \beta \langle x_1, -x_1 \rangle \\ &= \alpha \odot u \oplus \beta \odot u \\ \alpha \odot (u \oplus v) &= \alpha \langle x_1 + x_2, -x_1 - x_2 \rangle = \langle \alpha x_1 + \alpha x_2, -\alpha x_1 - \alpha x_2 \rangle \\ &= \langle \alpha x_1, -\alpha x_1 \rangle + \langle \alpha x_2, -\alpha x_2 \rangle \\ &= \alpha \langle x_1, -x_1 \rangle + \alpha \langle x_2, -x_2 \rangle \\ &= \alpha \odot u \oplus \alpha \odot v \end{align*}
Thus, S2, S4, and S5 are all satisfied. Because all ten axioms for a vector space are met, the set \(\mathbb {S} = \{ \langle x, y \rangle \in \mathbb {R}^2 \mid x + y = 0 \}\) is a vector space.
Exercise 0.1.3
Consider vectors in the form \(\langle x, y, z \rangle \) where the vector addition is defined as \(\langle x, y, z \rangle \oplus \langle x', y', z' \rangle = \langle x + x', y + y', z + z' \rangle \) and scalar multiplication as \(\alpha \langle x, y, z \rangle = \langle x^\alpha , y^\alpha , z^\alpha \rangle \). Identify if the vectors form a vector space.
Solution.
First, notice that the vector addition adheres to the standard vector addition. Therefore, we only have to look at scalar multiplication if any axioms are violated. From a glance, S1, S2, and S3 appears to be satisfied. Therefore, let’s take a look at S4. \[ (\alpha + \beta ) \odot u = \left ( x^{\alpha + \beta }, y^{\alpha + \beta }, z^{\alpha + \beta } \right ) \] Notice that in general, \(x^{\alpha + \beta } \neq x^\alpha + x^\beta \). Therefore, the set of vectors does not form a vector space.
Just as we have vector spaces \(\mathbb {R}^n\), we could also define spaces depending on the objects.
Definition 0.1.4
The symbol \(\mathbb {P}^n\) denotes the vector space composed of polynomials of degree at most \(n\) and adheres to the ten properties. \(\mathbb {M}_{m \times n}\) represents the vector space with matrices of order \(m \times n\) that adheres to the ten properties.
Therefore, we can say that \(\mathbb {R}^n\) is more or less the same as \(\mathbb {M}_{1 \times n}\). With the understandings above, we can derive some important theorems. Although most of them are intuitive, it’s always nice to formally prove them so let’s do that now.
Proof.
For the sake of contradiction, let \(0_1, 0_2 \in V\) be distinct additive identity element for a vector space \(V\). By definition, the following equations hold. \begin{align*} 0_1 \oplus 0_2 &= 0_1 \\ 0_1 \oplus 0_2 &= 0_2 \\ \therefore 0_1 \oplus 0_2 &= 0_1 = 0_2 \end{align*}
Because \(0_1 = 0_2\) is contradictory from the fact that \(0_1\) and \(0_2\) are distinct, the initial assumption that there exists distinct additive identity element in a vector space is false.
Continuing from the uniqueness of the zero vector in a vector space, we can assert the following theorem.
Theorem 0.1.6
For all \(u \in V\) for any vector space \(V\), \(v \in V\) such that \(u \oplus v = 0\) is unique.
Proof.
For the sake of contradiction, assume \(v, w \in V\) are distinct additive inverses of \(u\). By definition and properties of vector spaces, the following equations hold. \[ v = v \oplus 0 = v \oplus (u \oplus w) = (v \oplus u) \oplus w = 0 \oplus w = w \] Because \(v = w\) contradicts the definition that \(v\) and \(w\) are unique, the initial assumption that there exists more than one additive inverse of a vector in vector spaces is false.
Just as \(x = 0\) if \(x + x = x\), we could do similar things in vector spaces.
Theorem 0.1.7
For all \(v \in V\) and any vector space \(V\), \(v \oplus v = v\) if and only if \(v = 0\).
Proof.
First, the first half of the statement can be proven. Consider the following equations. \begin{align*} (v \oplus v) \oplus (-v) &= v \oplus (-v) = 0 \\ &= v \oplus (v \oplus (-v)) = v \oplus 0 = v \end{align*}
Therefore, if \(v \oplus v = v\), then \(v = 0\). Moreover, if \(v = 0\), then \(v \oplus v = 0 \oplus 0 = 0\). Therefore, the premise holds.
Now let’s take a look at theorems for scalar multiplications.
Theorem 0.1.8
The following properties hold for all vector space \(V\) over \(\mathbb {F}\), \(u, \mathbf {0} \in V\), and any scalars \(\alpha \).
- 1.
- \(0 \odot u = \mathbf {0}\)
- 2.
- \(\alpha \odot \mathbf {0} = \mathbf {0}\).
- 3.
- \((-1) \odot u = -u\)
- 4.
- \(u = -(-u)\)
- 5.
- \(\alpha \odot u = \mathbf {0}\) if and only if \(\alpha = 0\) or \(u = \mathbf {0}\).
Let’s start with the first proof.
Proof.
Consider the following equation. \[ 0 \odot u = (0 + 0) \odot u = 0 \odot u \oplus 0 \odot u \] By Theorem 0.1.7 , \(0 \odot u = \mathbf {0}\).
Below is the proof for the second property.
Proof.
Consider the following equation. \[ \alpha \odot \mathbf {0} = \alpha \odot (\mathbf {0} \oplus \mathbf {0}) = \alpha \odot \mathbf {0} \oplus \alpha \odot \mathbf {0} \] By Theorem 0.1.7 , \(\alpha \odot \mathbf {0} = \mathbf {0}\)
Continuing, here’s the proof for the third one.
Proof.
By the properties of vector spaces, \(\mathbf {0} = 0 \odot u = (1 - 1) \odot u = 1 \odot u \oplus (-1) \odot u = u \oplus (-1) \odot u \). Because \((-1) \odot u\) is the additive inverse of \(u\), \((-1) \odot u = -u\).
Below is the proof for the fourth one.
Proof.
Notice that \(-(-u)\) is the additive inverse of \(-u\). However, by definition, \(u\) is the additive inverse of \(-u\). By Theorem 0.1.6 , \(u = -(-u)\).
Finally, let’s prove the last property.
Proof.
It suffices to show the latter half of the statement as the first half is shown by the first two properties. Consider the following equations for \(\alpha \neq 0\). \begin{align*} \frac {1}{\alpha } \odot (\alpha \odot u) &= \frac {1}{\alpha } \odot \mathbf {0} = \mathbf {0} \\ &= \left ( \frac {1}{\alpha } \cdot \alpha \right ) \odot u = 1 \odot u = u \end{align*}
Therefore, if \(\alpha \neq 0\), then \(u = \mathbf {0}\). Similarly, if \(u \neq \mathbf {0}\), then \(\alpha = 0\) as the product of nonzero numbers is a nonzero number.
We have covered the fundamental concepts and properties in vector spaces. Before we move on with the next topic in vector spaces, I wanted to discuss briefly on dot product as it will frequently appear when you deal with vectors.
Definition 0.1.9
Dot product is an algebraic operation defined as the following for vectors \(a = [ a_1, \ldots , a_n]\) and \(b = [ b_1, \ldots , b_n]\) \[ a \cdot b = \sum _{k = 1}^n a_k b_k \]
Here notice that the number of elements of the vectors in dot product must be equal for the product to be defined. Below are some examples of dot products. \begin{align*} \begin {bmatrix} 1 \\ 2 \\ 3 \end {bmatrix} \cdot \begin {bmatrix} -1 \\ -2 \\ -3 \end {bmatrix} &= 1(-1) + 2(-2) + 3(-3) = -14 \\ \begin {bmatrix} 0 & 1 & 2 \end {bmatrix} \cdot \begin {bmatrix} 0 & 1 & 0 \end {bmatrix} &= 0 \cdot 0 + 1 \cdot 1 + 2 \cdot 0 = 1 \end{align*}
Note that this is very different from the matrix multiplication that we will discuss in the next note and we must write the notation \(\cdot \) to ensure that we are not doing matrix multiplication. With that in mind, let’s continue our discussion with subspaces.