1.1. Vector Spaces

Through this book, we use the symbol \(\F\)(F for field) to denote \(\R\) or \(\C\). Many mathematical definitions arises from simple concrete objects with certain properties. The abstraction of vector spaces comes from the nice properties possessed by \(\F^n\).

Definition 1.1

Let \(V\) be a set. We say \(V\) is a vector space over field \(F\)(or we simply say \(V\) is a vector space if \(\F\) is clear from the context) if the following properties hold:

1. Commutativity \(\mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u}\) for all \(\mathbf{u}, \mathbf{v} \in V \).

2. Associativity \((\mathbf{u}+\mathbf{v})+\mathbf{w} = \mathbf{u}+(\mathbf{v}+\mathbf{w})\) for all \(\mathbf{u}, \mathbf{v}, \mathbf{w} \in V\).

3. Additive Identity There exists an element \(\mathbf{0} \in V\) such that \(\mathbf{v} + \mathbf{0} = \mathbf{v}\) for all \(\mathbf{v} \in V\). (We only assume the existence of element \(\mathbf{0}\), nothing is said about its uniqueness. Although it is indeed unique, we still need to prove this.)

4. Additive Inverse For every \(\mathbf{v} \in V\), there exists an element \(\mathbf{w} \in V\) such that \(\mathbf{v} + \mathbf{w} = \mathbf{0}\) where \(\mathbf{0}\) is the same as the one given above.

5. Multiplicative Identity \(1 \mathbf{v} = \mathbf{v}\) for all \(\mathbf{v} \in V\) where \(1\) is simply the real number \(1\).

6. Distributive Property of Scalar Multiplication over Vector Addition \(a(\mathbf{u} + \mathbf{v}) = a\mathbf{u} + a\mathbf{v}\) for all \(a \in \F\) and all \(\mathbf{u}, \mathbf{v} \in V\).

7. Distributive Property of Scaler Multiplication over Scalar Addition \((a+b) \mathbf{v} = a\mathbf{v} + b\mathbf{v}\) for all \(a, b \in F\) and all \(\mathbf{v} \in V\).

Note

In fact, if a set satisfies property 1 - 4, we say that it is an Abelian group

As noted, we need to prove that such element \(\mathbf{0}\) is unique.

Proposition 1.1

Let \(V\) be an Abelian group. Suppose both elements \(\mathbf{0}\) and \(\mathbf{0}^\prime\) satisfy properties 3 in Definition 1.1. Then \(\mathbf{0} = \mathbf{0}^\prime\).

Proof. Regarding \(\mathbf{0}\) as an additive identity, by property 3, we have

(1.1)\[\mathbf{0}^\prime + \mathbf{0} = \mathbf{0}^\prime\]

On the other hand, regarding \(\mathbf{0}^\prime\) as an additive identity, we also have

(1.2)\[\mathbf{0} + \mathbf{0}^\prime = \mathbf{0}\]

Because the vector addition is commutative (property 1), the left-hand sides of (1.1) and (1.2) are equal and hence \(\mathbf{0} = \mathbf{0}^\prime\).

Now, we can safely say \(\mathbf{0}\) is the additive identity. We also say \(\mathbf{0}\) is the zero vector.

The next proposition is known as the cancellation property .

Proposition 1.2 (Cancellation Property)

Let \(V\) be an Abelian group. Suppose \(\mathbf{u}, \mathbf{v}, \mathbf{w} \in V\), we have

\[\mathbf{u} + \mathbf{w} = \mathbf{v} + \mathbf{w}\implies\mathbf{u} = \mathbf{v}\]

Proof. By property 4 in Definition 1.1, there exists \(\mathbf{w}^\prime\) such that \(\mathbf{w} + \mathbf{w}^\prime = \mathbf{0}\). It then follows that

\[\begin{split}& \quad& & \mathbf{u} + \mathbf{w} = \mathbf{v} + \mathbf{w}\\& \implies& & \mathbf{u} + (\mathbf{w}+ \mathbf{w}^\prime) = \mathbf{v} + (\mathbf{w} + \mathbf{w}^\prime) \\& \implies& & \mathbf{u} + \mathbf{0} = \mathbf{v} + \mathbf{0}\\& \implies& & \mathbf{u} = \mathbf{v}\end{split}\]

The number zero \(0\) times any vector is the zero vector \(\mathbf{0}\).

Proposition 1.3

Let \(V\) be a vector space. Then for any \(\mathbf{v} \in V\), we have \(0 \cdot \mathbf{v} = \mathbf{0}\).

The proof is simple. But we do not skip any intermediate steps to demonstrate how each property is applied.

Proof. Let \(\mathbf{v} \in V\) be arbitrary. We have

\[\mathbf{v} + 0 \cdot\mathbf{v} = 1 \cdot\mathbf{v} + 0 \cdot\mathbf{v} = (1 + 0) \cdot\mathbf{v} = 1 \cdot\mathbf{v} = \mathbf{v} = \mathbf{v} + \mathbf{0}\]

Then applying the cancellation property yields \(0 \cdot \mathbf{v} = \mathbf{0}\).

Any scalar in \(\F\) times the zero vector \(\mathbf{0}\) is the zero vector itself.

Proposition 1.4

Let \(V\) be a vector space. For any \(a \in \F\), we have \(a \mathbf{0} = \mathbf{0}\).

Proof. We have

\[a\mathbf{0} + \mathbf{0} = a \mathbf{0} = a (\mathbf{0} + \mathbf{0}) = a \mathbf{0} + a \mathbf{0}\]

Then by the cancellation property, we conclude that \(a\mathbf{0} = \mathbf{0}\).

The element \(\mathbf{w}\) in property 4 is also unique as we shall prove in the following proposition.

Proposition 1.5

Let \(V\) be an Abelian group. Pick an element \(\mathbf{v} \in V\). Suppose both elements \(\mathbf{w}\) and \(\mathbf{w}^\prime\) satisfy properties 4 in Definition 1.1. Then \(\mathbf{w} = \mathbf{w}^\prime\).

Proof. We have

\[\mathbf{w}^\prime + \mathbf{0} = \mathbf{w}^\prime\]

Replacing \(\mathbf{0}\) with \(\mathbf{v} + \mathbf{w}\) yields

\[\begin{split}& & & \mathbf{w}^\prime + (\mathbf{v} + \mathbf{w}) = \mathbf{w}\\& \implies& & (\mathbf{w}^\prime + \mathbf{v}) + \mathbf{w} = \mathbf{w}^\prime\\& \implies& & \mathbf{0} + \mathbf{w} = \mathbf{w}^\prime\\& \implies& & \mathbf{w} = \mathbf{w}^\prime\end{split}\]

where the last equality follows from Proposition 1.3.

Therefore, for each \(\mathbf{v}\), the choice of its additive inverse \(\mathbf{w}\) is unique. We may then say that such \(\mathbf{w}\) is the additive inverse of \(\mathbf{v}\). And to make the notation more intuitive, we shall denote the additive inverse of \(\mathbf{v}\) by \(-\mathbf{v}\).

The additive inverse \(-\mathbf{v}\) of \(\mathbf{v}\) can be computed by the scalar multiplication \(-1 \cdot \mathbf{v}\).

Proposition 1.6

Let \(V\) be a vector space and \(\mathbf{v} \in V\). We have \(-\mathbf{v} = -1 \cdot \mathbf{v}\).

Proof. We have

\[\mathbf{v} + (-1 \cdot\mathbf{v}) = 1 \cdot\mathbf{v} + (-1) \cdot\mathbf{v} = (1 + (-1)) \mathbf{v} = 0 \cdot\mathbf{v} = \mathbf{0}\]

This shows \(-1 \cdot \mathbf{v}\) is indeed the additive inverse of \(\mathbf{v}\).

The subtraction operator \(-: V \times V \to V\) is defined as \(\mathbf{u} - \mathbf{v} := \mathbf{u} + (-\mathbf{v})\).