Vector Spaces

1. Definition

First, we must give the definition of a vector space, although we will not use this definition very often in what follows.

A Vector Space is a set $V$ together with a rule for adding two elements (or vectors) $x, y \in V$ and a rule for scaling any element (or vector) $x \in V$ by a real number $\alpha \in \mathbb{R},$ such that the following axioms hold for all $x, y, z \in V$ and every $\alpha, \beta \in \mathbb{R}:$

$x + y = y + x$ (addition is commutative)
$x + ( y + z ) = ( x + y ) + z$ (addition is associative)
There is a unique element in $V$ called zero, often denoted $\vec{0},$ such that $x + \vec{0} = x$ for all $x \in V$
Each element $x \in V$ has a unique additive inverse $-x \in V$ such that $x + (-x) = \vec{0}$
$1 \cdot x = x$ for all $x \in V$
$(\alpha \beta) x = \alpha (\beta x)$ (scaling is associative)
$\alpha (x + y) = \alpha x + \alpha y$ (distribution of scalars)
$(\alpha + \beta) x = \alpha x + \beta x$ (distribution of vectors)

2. Examples

The best way to understand the notion of a vector space is to see some examples:

$n$-space: $\mathbb{R}^n$ for $n = 1, 2, \dots$

Matrices:

$$ M_{r \, \times \, c}(\mathbb{R}) = \left\{ \quad \left[ \begin{array}{cccc} a_{1,1} & a_{1,2} & \cdots & a_{1,c} \\ a_{2,1} & a_{2,2} & \cdots & a_{2,c} \\ \vdots & \vdots & \ddots & \vdots \\ a_{r,1} & a_{r,2} & \cdots & a_{r,c} \\ \end{array} \right] \quad \middle\vert \quad \begin{array}{c} a_{\,i,\,j} \in \mathbb{R} \\ \textrm{for } i = 1, 2, \dots , r \\ \textrm{and } j = 1, 2, \dots , c \\ \end{array} \quad \right\} $$

The notation above describes matrices with $r$ rows and $c$ columns. If a matrix has the same number of rows and columns, it is referred to as a square matrix. Square matrices use slightly different notation - instead of $M_{n \times n}(\mathbb{R})$ we will simply write $M_n(\mathbb{R})$.

Polynomials:

$$ \mathcal{P}_n = \left\{ a_0 + a_1 t + a_2 t^2 + \cdots + a_n t^n \, \middle\vert \, \, a_0, a_1, \dots, a_n \in \mathbb{R} \right\} $$

For each $n$, we have $\mathcal{P}_n$, polynomials with real coefficients of degree less than or equal to $n$. (Note: the set of polynomials with degree exactly $n$ is not a vector space.) These vector spaces are nested, for example $\mathcal{P}_1 \subseteq \mathcal{P}_2$, and all of them lie in the much larger vector space of all polynomials,

$$ \mathcal{P} = \left\{ a_0 + a_1 t + a_2 t^2 + \cdots + a_n t^n \, \middle\vert \, \, n = 0, 1, 2, \dots \textrm{ and } a_0, a_1, \dots, a_n \in \mathbb{R} \right\} $$

so that we have

$$\mathcal{P}_0 \subseteq \mathcal{P}_1 \subseteq \mathcal{P}_2 \subseteq \cdots \subseteq \mathcal{P}_n \subseteq \cdots \subseteq \mathcal{P}$$
Complex Numbers:

$$ \mathbb{C} = \left\{ \alpha + \beta i \, \middle\vert \, \, \alpha, \beta \in \mathbb{R} \right\} $$

This is a vector space under addition and scaling by real numbers. Notice that we have the notion of multiplication and division (and thus multiplicative inverses) in this set, but those things do not play into the definition of $\mathbb{C}$ as a vector space. Those extra rules give $\mathbb{C}$ more structure than other vector spaces (and those structures have names - $\mathbb{C}$ is also a commutative ring and a field), but when we ignore that extra structure we see that $\mathbb{C}$ is a vector space.

Real-Valued Functions:

$$ C(\mathbb{R}) = \left\{ \, f : \mathbb{R} \to \mathbb{R} \, \middle\vert \, \, f \textrm{ is continuous on } \mathbb{R} \right\} $$

Again, we can multiply functions (making $C(\mathbb{R})$ an object called a commutative ring), but if we ignore that then we can see that $C(\mathbb{R})$ is a vector space. Often, $C(\mathbb{R})$ is denoted $C^{\, 0}(\mathbb{R})$, and in general

$$ C^{\, k}(\mathbb{R}) = \left\{ \, f : \mathbb{R} \to \mathbb{R} \, \middle\vert \, \, \frac{d^kf}{dx^k} = f^{(k)} \textrm{ is continuous on } \mathbb{R} \right\} $$

and

$$ C^{\, \infty}(\mathbb{R}) = \left\{ \, f : \mathbb{R} \to \mathbb{R} \, \middle\vert \, \, f \textrm{ is infinitely differentiable } \right\} $$

These vector spaces are nested again - if $f$ has a continuous second derivative, then $f$ has a continuous first derivative. We get

$$C^{\, \infty}(\mathbb{R}) \subseteq \cdots \subseteq C^{\, k}(\mathbb{R}) \subseteq \cdots \subseteq C^{\, 2}(\mathbb{R}) \subseteq C^{\, 1}(\mathbb{R}) \subseteq C^{\, 0}(\mathbb{R}) = C(\mathbb{R})$$

In fact, since polynomials, $\mathcal{P}$, are infinitely differentiable functions on $\mathbb{R}$, we have

$$\mathcal{P}_0 \subseteq \mathcal{P}_1 \subseteq \mathcal{P}_2 \subseteq \cdots \subseteq \mathcal{P}_n \subseteq \cdots \subseteq \mathcal{P} \subseteq C^{\, \infty}(\mathbb{R}) \subseteq \cdots $$ $$\cdots \subseteq C^{\, k}(\mathbb{R}) \subseteq \cdots \subseteq C^{\, 2}(\mathbb{R}) \subseteq C^{\, 1}(\mathbb{R}) \subseteq C^{\, 0}(\mathbb{R}) = C(\mathbb{R})$$

Solutions to homogeneous O.D.E.'s:
Let $$L[y] = a_n(t) \, y^{(n)} + a_{n-1}(t) \, y^{(n-1)} + \cdots + a_2(t) \, y^{\, \prime \prime} + a_1(t) \, y^{\, \prime} + a_0(t) \, y = 0$$ be an $n$th order, linear, homogeneous differential equation. The solutions form a vector space:

$$ V_L = \left\{ \, y \in C(\mathbb{R}) \, \middle\vert \, \, L[y] = 0 \right\}$$

Note that $V_L$ is always a subset of $C(\mathbb{R})$, although it may lie in a smaller subset. For example, when $L$ is order $n$, $V_L \subseteq C^{\, n-1}(\mathbb{R})$. When $L$ has constant coefficients (i.e. when $a_n(t), \dots , a_0(t) \in \mathbb{R}$), then $V_L \subseteq C^{\, \infty}(\mathbb{R})$. Can you think of an example, $L$, for which $V_L \subseteq \mathcal{P}$?

Vector-Valued Functions:
A vector valued function is a function $F: \mathbb{R} \to \mathbb{R}^n$. It can be expressed as a tuple of functions: $$F(t)=\left[ \begin{array}{c} \, \, f_1(t) \, \, \\ f_2(t) \\ \vdots \\ f_n(t) \end{array} \right]$$ where each $f_i$ is a real-valued function. The function may also be written as $F(t) = \left[ \, \, f_1(t)\, , \, f_2(t)\, , \dots, \, f_n(t) \, \right]$. The graph of a vector-valued function is a curve in $\mathbb{R}^n$. For example, $F(t) = \left[ \, \cos(t)\, , \, \sin(t) \, \right]$ has graph the unit circle in the plane and $G(t) = \left[ \, t \cos(t)\, , \, t \sin(t) \, \right]$ has a spiral in the plane as its graph. Notice that this graph is not like the graph of a function $y = f(t)$ since we can not see the $t$-axis. One can visualize this axis by watching the curve being graphed over time, $t$. Click play in the left-hand column of the following link to see the curve being graphed over time:

https://www.desmos.com/calculator/zd3nem7fhu
Another example is the spiral in $3$-space, $F(t) = \left[ \, \cos(t)\, , \, \sin(t) \, , \, t \, \right]$, shown here in the Sage cell:

Just like real-valued functions, we can add and scale vector-valued functions by adding and scaling their images pointwise in the codomain, $\mathbb{R}^n$. Thus we have new vector spaces

$$ C^0(\mathbb{R},\mathbb{R}^n) = C(\mathbb{R},\mathbb{R}^n) = \left\{ \, \, \left[ \begin{array}{c} \, \, f_1 \, \, \\ f_2 \\ \vdots \\ f_n \end{array} \right] : \mathbb{R} \to \mathbb{R}^n \, \middle\vert \, \, \begin{array}{c} f_i \textrm{ is continuous on } \mathbb{R} \\ \textrm{ for } i = 1, 2, \dots , n \\ \end{array} \right\} $$
$$ C^{\, k}(\mathbb{R},\mathbb{R}^n) = \left\{ \, \, \left[ \begin{array}{c} \, \, f_1 \, \, \\ f_2 \\ \vdots \\ f_n \end{array} \right] : \mathbb{R} \to \mathbb{R}^n \, \middle\vert \, \, \begin{array}{c} \frac{d^kf_i}{dx^k} = f_i^{(k)} \textrm{ is continuous on } \mathbb{R} \\ \textrm{ for } i = 1, 2, \dots , n \\ \end{array} \right\} $$
$$ C^{\, \infty}(\mathbb{R},\mathbb{R}^n) = \left\{ \, \, \left[ \begin{array}{c} \, \, f_1 \, \, \\ f_2 \\ \vdots \\ f_n \end{array} \right] : \mathbb{R} \to \mathbb{R}^n \, \middle\vert \, \, \begin{array}{c} f_i \textrm{ is infinitely differentiable } \\ \textrm{ for } i = 1, 2, \dots , n \\ \end{array} \right\} $$

Note: For solving linear systems of differential equations, we will focus solely on $C^{\, \infty}(\mathbb{R},\mathbb{R}^n)$, and following our text we will write these vector-valued functions as $$\vec{x}(t) = \left[ \begin{array}{c} \, \, x_1(t) \, \, \\ x_2(t) \\ \vdots \\ x_n(t) \end{array} \right].$$

Solutions to homogeneous, first order systems of O.D.E.'s:
Given $A \in M_n(\mathbb{R})$, consider the homogeneous, first order system of differential equations given by $$\frac{d}{dt} \vec{x} = A \vec{x}.$$ A solution to this equation is a vector-valued function $\vec{x}(t) \in C^{\, \infty}(\mathbb{R},\mathbb{R}^n)$. The set of all solutions forms a vector space

$$ V_A = \left\{ \, \vec{x} \in C^{\, \infty}(\mathbb{R},\mathbb{R}^n) \, \middle\vert \, \, \frac{d}{dt} \vec{x} = A \vec{x} \right\}$$

Note: Homogeneous O.D.E.'s $L$ have solution set $V_L \subseteq C(\mathbb{R})$ and linear systems given by a square matrix $A \in M_n(\mathbb{R})$ have solution set $V_A \subseteq C^{\, \infty}(\mathbb{R},\mathbb{R}^n)$.

A note on domain:
Our definition of $C^{\, n}(\mathbb{R})$ has one flaw: it only allows for functions defined on the entire real number line. For example, $\ln (t)$, $\arcsin(t)$, $\sqrt{t}$, and $1\,/\,t$ are not members of this vector space. A simple generalization is to consider a subset $U \subseteq \mathbb{R}$ and define the vector spaces

$ C(U) = \left\{ \, f : U \to \mathbb{R} \, \middle\vert \, \, f \textrm{ is continuous on } U \right\} $
$$ C^{\, k}(U) = \left\{ \, f : U \to \mathbb{R} \, \middle\vert \, \, \frac{d^kf}{dx^k} = f^{(k)} \textrm{ is continuous on } U \right\} $$
$ C^{\, \infty}(U) = \left\{ \, f : U \to \mathbb{R} \, \middle\vert \, \, f \textrm{ is infinitely differentiable } \right\} $

Then we can easily account for domains:

$\ln(t) \in C((0,\infty))$
$\arcsin(t) \in C([-1,1])$
$\sqrt{t} \in C([0,\infty))$
$1\,/\,t \in C((-\infty,0)\cup(0,\infty))$

This is especially important for differential equations. For example, the Euler Equation $$L[\,y\,] = 2\, t^2 \, y^{\, \prime \prime} + 3\, t \, y^{\, \prime} - y = 0$$ has fundamental solutions $y_1(t) = \sqrt{t}$ and $y_2(t) = 1\,/\,t$. Thus, in this case, $V_L = \left\{ c_1 \, \sqrt{t} + c_2 \, \left(\,1\,/\,t\,\right) \, \, \middle\vert \, \, c_1, c_2 \in \mathbb{R} \right\} \subseteq C\left((0,\infty)\right)$.

3. Geometric Intuition

Every vector space has some dimension. The line $\mathbb{R}^1$ is $1$ dimensional, the plane $\mathbb{R}^2$ is $2$ dimensional, and the $3$-space $\mathbb{R}^3$ is $3$ dimensional. We'll define dimension formally later, but for now I'm relying on your intuition based on these three examples. We can guess from these examples that the dimension of $\mathbb{R}^n$ is $n$.

How can we determine the dimensions of other vector spaces? Another way to describe dimension is to say that every vector space "acts like" $\mathbb{R}^n$ for some $n,$ and then $n$ is the dimension. The term for "acts like" is isomorphic and the symbol is $\cong$. For example, $$M_2(\mathbb{R}) = \left\{ \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] \, \Bigg\vert \, \, a, b, c, d \in \mathbb{R} \right\}$$ has four coordinates and thus it "acts like" $\mathbb{R}^4$, we write $M_2(\mathbb{R}) \cong \mathbb{R}^4$, and we say $M_2(\mathbb{R})$ is $4$ dimensional. Here are some more examples:

$\mathbb{C} \cong \mathbb{R}^2$
$M_{\, r \, \times \, c \, }(\mathbb{R}) \cong \mathbb{R}^{r \cdot c}$
$M_n(\mathbb{R}) \cong \mathbb{R}^{n^2}$
$\mathcal{P}_n \cong \mathbb{R}^{n+1}$

Some vector spaces are so big they are infinite dimensional. We will use the symbol $\mathbb{R}^{\infty}$ to denote these:

$\mathcal{P} \cong \mathbb{R}^{\infty}$
$C^{\, \infty}(\mathbb{R}) \cong \mathbb{R}^{\infty}$
$C^{\, k}(\mathbb{R}) \cong \mathbb{R}^{\infty}$
$C^{\, \infty}(\mathbb{R},\mathbb{R}^n) \cong \mathbb{R}^{\infty}$
$C^{\, k}(\mathbb{R},\mathbb{R}^n) \cong \mathbb{R}^{\infty}$

From our text we know that if $L$ is an $n$th order, homogeneous O.D.E. with constant coefficients, than its solution set $V_L \cong \mathbb{R}^n$ is $n$-dimensional. We also know that if $A \in M_n(\mathbb{R})$, then the solution set to the system $\displaystyle\frac{d}{dt}\vec{x} = A \vec{x}$, $V_A \cong \mathbb{R}^n$, is $n$-dimensional.

If $L$ is order $n$, then $V_L \cong \mathbb{R}^n$
If $A \in M_n(\mathbb{R})$, then $V_A \cong \mathbb{R}^n$