Vector spaces

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We have defined addition and scalar multiplication of vectors. These operations satisfy a set of laws that characterize vector algebra. Conversely, any set equipped with ``addition'' and ``scalar multiplication'' that satisfy the laws of vector algebra is considered a vector space. In this view, vectors are elements of a vector space.



We usually assume that vectors are elements of \(\mathbb{R}^n\) for some \(n\in \mathbb{N}\). This \(\mathbb{R}^n\) is an example of a vector space.

Remark. In mathematics, in general, we tend to use the word ``space'' to denote a set with some additional ``structures''. e.g., vector space, probability space, topological space, Banach space, inner product space, Hilbert space, etc. □

The following results apply to any vector spaces (e.g. \(\mathbb{C}^n\)). Therefore we use \(V\) to represent an arbitrary vector space. We use \(K\) to represent the field underlying the vector space. We say \(V\) is a vector space over the field \(K\).

Example. 

  • \(\mathbb{R}^n\) is a vector space over the field \(\mathbb{R}\).
  • \(\mathbb{C}^n\) is a vector space over the field \(\mathbb{C}\).
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The following lemma is a list of algebraic laws for vectors. We use boldface letters such as \(\mathbf{u}, \mathbf{v}\) to represent vectors and Greek letters such as \(\lambda, \mu,\cdots\) to represent scalars. 

Lemma 

  1. For all \(\mathbf{u}, \mathbf{v}\in V\), \(\mathbf{u} + \mathbf{v} \in V\).  (Vector addition is closed in \(V\).)
  2. For all \(\mathbf{u}, \mathbf{v}\in V\), \(\mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u}\). (Vector addition is commutative.)
  3. For any \(\mathbf{u}, \mathbf{v}, \mathbf{w} \in V\), \((\mathbf{u}  + \mathbf{v}) + \mathbf{w} = \mathbf{u} + (\mathbf{v} + \mathbf{w})\). (Vector addition is associative.)
  4. There exists an element \(\mathbf{0}\in V\) such that for all \(\mathbf{u}\in V\), \(\mathbf{0} + \mathbf{u} = \mathbf{u}\). (The additive identity element exists and is called \(\mathbf{0}\) (zero).)
  5. For each \(\mathbf{u}\in V\), there exists \(\mathbf{-u}\) such that \(\mathbf{u} + (\mathbf{-u}) = \mathbf{0}\). (The additive inverse (\(-\mathbf{u}\)) always exists.)
  6. For each \(\lambda \in K\), and for each \(\mathbf{u}\in V\), \(\lambda\mathbf{u} \in V\). (Multiplication by a scalar is closed in \(V\).)
  7. For each \(\mathbf{u}\in V\), for each \(\lambda, \mu\in K\), \((\lambda + \mu)\mathbf{u} = \lambda\mathbf{u} + \mu\mathbf{u}\). (The distributive law: multiplication by a vector distributes over scalar addition. [Note: Scalars are elements of the field \(K\) so that addition and multiplication are defined.])
  8. For all \(\mathbf{u}, \mathbf{v} \in V\), for all \(\lambda \in K\), \(\lambda(\mathbf{u} + \mathbf{v}) = \lambda\mathbf{u} + \lambda\mathbf{v}\).(The distributive law: multiplication by a scalar distributes over vector addition.)
  9. For all \(\mathbf{u}\in V\), \(1\cdot \mathbf{u} = \mathbf{u}\). (The multiplicative identity of \(K\) acts as the multiplicative identity for the scalar multiplication in \(V\).)
  10. For all \(\lambda, \mu \in K\), for all \(\mathbf{u} \in V\), \((\lambda \mu)\mathbf{u} = \lambda(\mu\mathbf{u})\). (Scalar multiplication is associative.)
Example (\(\mathbb{R}^2\)). Let's prove some of the above laws for \(\mathbb{R}^2\). Let \(\mathbf{u} = (u_1, u_2)\) and \(\mathbf{v} = (v_1, v_2)\) and \(\lambda \in \mathbb{R}\). 
(Law 1) Recall that vector addition is defined by
\[\mathbf{u} + \mathbf{v} = (u_1, u_2) + (v_1, v_2) = (u_1 + v_1, u_2 + v_2).\]
Clearly, 
\[(u_1 + v_1, u_2 + v_2) \in \mathbb{R}^2.\]
Therefore, \(\mathbb{R}^2\) is closed under vector addition.
(Law 6) Scalar multiplication is defined by
\[\lambda\mathbf{u} = \lambda(u_1, u_2) = (\lambda u_1, \lambda u_2),\]
and clearly,
\[(\lambda u_1, \lambda u_2)\in\mathbb{R}^2.\]
Thus, \(\mathbb{R}^2\) is closed under scalar multiplication.
(Law 4) The zero element \(\mathbf{0}\) is given by \(\mathbf{0} = (0, 0)\) where \(0\) is the real number zero.
In fact,
\[\mathbf{0} + \mathbf{u} = (0, 0) + (u_1, u_2) = (0+u_1, 0+u_2) = (u_1, u_2).\]
(Law 5) The additive inverse of \(\mathbf{u} = (u_1, u_2)\) is \(-\mathbf{u} = (-u_1, -u_2)\). In fact,
\[\mathbf{u} + (-\mathbf{u}) = (u_1, u_2) + (-u_1, -u_2) = (u_1 - u_1, u_2 - u_2) = (0,0).\]
(Law 9) The multiplicative identity for the scalar multiplication is \(1 \in \mathbb{R}\). In fact,
\[1\cdot(u_1, u_2) = (1\cdot u_1, 1\cdot u_2) = (u_1, u_2).\]

You should check other laws, too. □

Definition (Vector space)

Let \(V\) be a set equipped with addition and scalar multiplication, with scalars being elements of the field \(K\). If all of the above laws (1-10) are satisfied, we say that \(V\) is a vector space.

Although it may look similar, a vector space is not a field. For example, we don't have multiplication between vectors but only between scalars and vectors. A vector space is actually an algebraic structure of its own kind.
See also: Fields





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