Vector spaces

We previously described scalars and now we turn to vectors.

Definition

A $k$-vector space, or a vector space over $k$, is a set $V$ equipped two operations: \(\begin{aligned} + : V \times V & \to V \\ \cdot : k \times V & \to V \end{aligned}\) satisfying the following conditions:

• $+$ is a associative: $(v_1+v_2)+v_3 = v_1+(v_2+v_3)$.
• $+$ is commutative: $v_1 + v_2 = v_2 + v_1$.
• There is an element $0 \in V$ with $0 + v = v + 0 = 0$ for any $v \in V$.
• For any element $v \in V$, there is another $-v$ with $v + (-v) = (-v) + v = 0$.
• $\cdot$ distributes over (both) $+$: \(c \cdot (v_1 + v_2) = c\cdot v_1 + c \cdot v_2\) and \((c_1 + c_2) \cdot v = c_1 \cdot v + c_2 \cdot v\)
• $1 \cdot$ is the identity: \(1 \cdot v = v\) for all $v \in V$.
• Finally, $\times$ in $k$ and $\cdot$ have the following relation \((c_1 \times c_2) \cdot v = c_1 \cdot (c_2 \cdot v)\)

Elements of a vector space $V$ are called vectors.

The conditions on $+$ for $V$ are reminiscent of what we saw for fields.

However, $\cdot$ is a bit different. It takes two different inputs: one a scalar from $k$ and the other a vector from $V$. We can translate the identity \((c_1 \times c_2) \cdot v = c_1 \cdot (c_2 \cdot v)\) as saying: scaling by the product of $c_1$ and $c_2$ is the same as scaling by $c_2$ and then scaling the result by $c_1$.

Examples

• Any field is a vector space over itself. Indeed, we can take $\cdot = \times$. For the case of $k = \mathbb{R}$,

• For any field $k$, the set consisting of a single point is a vector space over $k$. Let’s suggestively call that point $0$. Then we set $0+0 = 0$

• For $k = \mathbb{R}$, the Euclidean spaces $\mathbb{R}^n$ are vector spaces. In general, for a field $k$, we have a vector space $k^n$ consisting of length $n$ lists of elements of $k$. Addition is \(\begin{pmatrix} v_1 \\ v_2 \\ \vdots \\ v_n \end{pmatrix} + \begin{pmatrix} w_1 \\ w_2 \\ \vdots \\ w_n \end{pmatrix} = \begin{pmatrix} v_1+w_1 \\ v_2+w_2 \\ \vdots \\ v_n+w_n \end{pmatrix}\) and scalar multiplication is given by \(c \begin{pmatrix} v_1 \\ v_2 \\ \vdots \\ v_n \end{pmatrix} = \begin{pmatrix} cv_1 \\ cv_2 \\ \vdots \\ cv_n \end{pmatrix}\) analogously to the case $k = \mathbb{R}$.

• Given an $m \times n$ matrix $A$ and vector $\mathbf{b} \in k^m$, the null space $\mathcal Z(A)$ is a $k$-vector space. We will use vector addition and scalar multiplication. To verify we indeed have a vector space, we need to check that:
  • if $\mathbf{v}, \mathbf{w} \in \mathcal Z(A)$ then $\mathbf{v} + \mathbf{w} \in \mathcal Z(A)$ and
  • if $\mathbf{v} \in \mathcal Z(A)$, then $c \mathbf{v} \in \mathcal Z(A)$

  We remember that $\mathbf{v} \in \mathcal Z(A)$ is another way to state that $A \mathbf{v} = \mathbf{0}$. So we can translate the previous statements into

  • if $A \mathbf{v} = \mathbf{0}$ $A\mathbf{w} = \mathbf{0}$ then $A(\mathbf{v} + \mathbf{w}) = \mathbf{0}$ and
  • if $A \mathbf{v} = \mathbf{0}$, then $A(c \mathbf{v}) = \mathbf{0}$

  Since matrix multiplication is distributive over addition, we have \(A(\mathbf{v} + \mathbf{w}) = A \mathbf{v} + A \mathbf{w} = \mathbf{0}\) if $A \mathbf{v}, A \mathbf{w} = \mathbf{0}$.

  Similarly, since $Ac = cA$ for a scalar $c$ and a matrix $A$, we have \(A(c\mathbf{v}) = c A(\mathbf{v}) = \mathbf{0}\) if $A \mathbf{v} = \mathbf{0}$.

  This example is a case of more general statement about subspaces.

• We can contrast the previous example with the case of $\mathcal Z(A \mid \mathbf{b})$ for $\mathbf{b} \neq \mathbf{0}$. Consider $x+y = 1$. Then, $(1 \ 0)^T$ is a solution but $2 (1 \ 0)^T = (2 \ 0)$ is not!

  Solutions to the homogeneous system $A \mathbf{x} = \mathbf{0}$, ie null spaces, are vector spaces but solutions to an inhomogeneous system are never vector spaces.

• Ranges are also vector spaces. For any matrix $A$, we know there is another matrix $B$ with $\mathcal R(A) = \mathcal Z(B)$. As we just saw, $\mathcal Z(B)$ is a vector space. We will see more directly that ranges are vector spaces when we talk about spans.

• Let $\operatorname{Mat}_{m,n}(k)$ be the set of $m \times n$ matrices with entries in the field $k$. This is a $k$-vector space with matrix addition and scalar multiplication.

• There are also more exotic examples. Let \(\operatorname{Fun}(\mathbb{R},\mathbb{R}) = \lbrace f: \mathbb{R} \to \mathbb{R} \rbrace\) be the set of all real-valued functions of one variable with domain all of $\mathbb{R}$. Some examples of elements include \(e^x, \ 5x^5 + 4x^4 + 3x^3 + 2x^2 + x + 1, \ \frac{1}{1+x^2}, \sin(x),\ldots\) This is a $\mathbb{R}$-vector space with the sum of functions $f+g$ given by \(x \mapsto f(x) + g(x)\) and scalar multiplication $cf$ by \(x \mapsto cf(x)\) The element $0$ is the constant function with value $0$.

Consequences of the definition

Some facts are true for a general vector space because they follow logically from the definition.

One consequence is that the element $-v$ that satisfies $v + (-v) = 0$ is uniquely determined by $v$. As such, the notation $-v$ (implicitly saying it only depends on $v$) is sensible.

Lemma. Let $V$ be a vector space and $v \in V$. If \(v + w = 0 = v + w^\prime\) then $w = w^\prime$.

We know the following facts.

Lemma. Let $V$ be a vector space. Then

• $0 \cdot v = 0$ for any $v \in V$
• $c \cdot 0 = 0$ for any $c \in k$, and
• $(-1) \cdot v = -v $ for any $v \in V$.