19.2 Vector spaces

19.2.1 The vector space axioms

There are eight axioms in total, but I find it easier to remember them this way:

Definition 19.2.1

A vector space is a set V over a field $\mathbb{K}$ (elements of which are called \sayscalars) equipped with an operator $+$ called \sayvector addition which is an operator taking two elements of $V$ and returning a single element in $V$ , and an operator $\cdot$ called \sayscalar multiplication which takes a scalar and a vector, and outputs a vector.

We have the following eight axioms,

1.

The first four axioms are equivalent to stating that $(\textsf{V},+)$ must be an Abelian group.
2.

We have two kinds of distributivity, one is that if $\mathbf{x},\mathbf{y}\in\textsf{V}$ and $a\in\mathbb{K}$ , then

$a\cdot(\mathbf{x}+\mathbf{y})=a\cdot\mathbf{x}+a\cdot\mathbf{y}$ (19.40)
3.

The second is that if $a,b\in\mathbb{K}$ and $\mathbf{x}\in\textsf{V}$ , then

$\mathbf{x}\cdot(a+b)=\mathbf{x}\cdot a+\mathbf{x}\cdot b$ (19.41)
4.

The neutral element (e.g. in $\mathbb{R}$ this is $1$ ) in $\mathbf{K}$ has the following property,

$\forall\mathbf{x}\in\textsf{V}\hskip 12.0pt1\cdot\mathbf{}{x}=\mathbf{x}$ (19.42)
5.

We also have a kind of \saymultiplicative distributivity

$a(b\mathbf{x})=(ab)\mathbf{x}$ (19.43)

Not exactly the most exciting stuff, but we can’t build castles without foundations! I’m not a structural engineer, but I’m pretty sure this is a true statement.

19.2.2 Linear independence

This is a $\text{very important}^{TM}$ concept in linear algebra.

Definition 19.2.2

Let $v_{1},v_{2},...,v_{n}$ be some vectors in a vector space V, and let $a_{1},a_{2},...,a_{n}$ be some scalars in the field $\mathbb{K}$ (over which this vector space is defined).

We say these vectors are linearly independent if and only if

a_{1}v_{1}+a_{2}v_{2}+...+a_{n}v_{n}=0\implies a_{1}=a_{2}=...=a_{n}=0.

(19.44)

In words, this means \sayif the only values for all the $a$ s which satisify $a_{1}v_{1}+a_{2}v_{2}+...+a_{n}v_{n}=0$ are when all the $a$ s are zero, then the vectors are linearly independent.

19.2.3 Bases

Change of basis

Let us suppose that we have two sets of basis vectors for the same vector space $V$ . It doesn’t really matter what we call them, but $\mathcal{B}$ and $\mathcal{C}$ are names as good as any. These vectors can be written in the form

	$\displaystyle\mathcal{B}=\{\beta_{1},\beta_{2},...,\beta_{\dim(V)}\}$		(19.45)
	$\displaystyle\mathcal{C}=\{\gamma_{1},\gamma_{2},...,\gamma_{\dim(V)}\}$		(19.46)

For any vector $\mathbf{v}\in V$ we can always write it in the co-ordinate system $\mathcal{B}$ by writing the vector as a linear combination⁵⁵ 5 This is always possible because $\mathbf{B}$ is a basis for $V$ . of the vectors in $\mathcal{B}$ . We can write this as

[\mathbf{v}]_{\mathcal{B}}=\begin{pmatrix}v_{1}\\ v_{2}\\ ...\\ v_{\dim(V)}\end{pmatrix}

(19.47)

Where $v_{1},v_{2},...,v_{\dim(V)}$ are such that

\mathbf{v}=v_{1}\beta_{1}+v_{2}\beta_{2}+...+v_{\dim(V)}\beta_{\dim(V)}

(19.48)

That is, they are the coefficients needed to write $\mathbf{v}$ as a linear combination of $\mathcal{B}$ . This also helps to understand why for example the vector space of $2\times 2$ symmetric matrices⁶⁶ 6 i.e. those in the form $\begin{pmatrix}a&b\\ b&c\end{pmatrix}$ (19.49) is three dimensional; we can write every matrix as a vector of dimension $3\times 1$ where each coefficient denotes what to multiply each basis vector by to obtain our specific vector.

But what if we want to find a way to translate $[\mathbf{v}]_{\mathcal{C}}$ into $[\mathbf{v}]_{\mathcal{B}}$ ? This is actually doable using a single matrix. Here’s how. We start by applying the definition of $[\mathbf{v}]_{\mathcal{B}}$ , that is, we have that $[\mathbf{v}]_{\mathcal{B}}=[v_{1},v_{2},...,v_{\dim(V)}]$ if and only if

\mathbf{v}=v_{1}\beta_{1}+v_{2}\beta_{2}+...+v_{\dim(V)}\beta_{\dim(V)}

(19.50)

To find $[\mathbf{v}]_{\mathcal{C}}$ , it is sufficient to find the $\mathbf{v}$ in terms of the basis vectors in $\mathbf{c}$ . How do we do this? A straightforward approach is to write every vector in $\mathcal{B}$ in terms of those in $\mathcal{C}$ and then to substitute for them, which removes all the $\mathcal{B}$ -vectors and means that we instead have $\mathcal{C}$ -vectors.

Because $\mathcal{B}$ and $\mathcal{C}$ are both basis for $V$ , we can write every vector in $\mathcal{V}$ in terms of those in $C$ .

	$\displaystyle\beta_{1}=\alpha_{1,1}\gamma_{1}+\alpha_{2,1}\gamma_{2}+...+% \alpha_{\dim(V),1}\gamma_{\dim(V)}$		(19.51)
	$\displaystyle\beta_{2}=\alpha_{1,2}\gamma_{1}+\alpha_{2,2}\gamma_{2}+...+% \alpha_{\dim(V),2}\gamma_{\dim(V)}$		(19.52)
	$\displaystyle...$		(19.53)
	$\displaystyle\beta_{\dim(V)}=\alpha_{1,\dim(V)}\gamma_{1}+\alpha_{2,\dim(V)}% \gamma_{2}+...+\alpha_{\dim(V),\dim(V)}\gamma_{\dim(V)}$		(19.54)

We can then substitute this into the linear combination of $\mathbf{v}$ in terms of the basis vectors in $\mathcal{B}$ , giving

$\displaystyle\mathbf{v}$	$\displaystyle=v_{1}\left(\alpha_{1,1}\gamma_{1}+\alpha_{2,1}\gamma_{2}+...+% \alpha_{\dim(V),1}\gamma_{\dim(V)}\right)$	(19.56)
	$\displaystyle\quad+v_{2}\left(\alpha_{1,2}\gamma_{1}+\alpha_{2,2}\gamma_{2}+..% .+\alpha_{\dim(V),2}\gamma_{\dim(V)}\right)$	(19.57)
	$\displaystyle\quad+...$	(19.58)
	$\displaystyle\quad+v_{\dim(V)}\left(\alpha_{1,\dim(V)}\gamma_{1}+\alpha_{2,% \dim(V)}\gamma_{2}+...+\alpha_{\dim(V),\dim(V)}\gamma_{\dim(V)}\right)$	(19.59)

This looks scary, but we just need to stick to the definitions and keep our goal in mind; writing $\mathbf{v}$ in terms of all the $\gamma$ . We can move things around to obtain

$\displaystyle\mathbf{v}$	$\displaystyle=\left(v_{1}\alpha_{1,1}+v_{2}\alpha_{2,1}+...+v_{\dim(V)}\alpha_% {\dim(V),1}\right)\gamma_{1}$	(19.60)
	$\displaystyle\quad+\left(v_{2}\alpha_{1,2}+v_{2}\alpha_{2,2}+...+v_{\dim(V)}% \alpha_{\dim(V),2}\right)\gamma_{2}$	(19.61)
	$\displaystyle\quad+...$	(19.62)
	$\displaystyle\quad+\left(v_{2}\alpha_{1,\dim(V)}+v_{2}\alpha_{2,\dim(V)}+...+v% _{\dim(V)}\alpha_{\dim(V),\dim(V)}\right)\gamma_{\dim(V)}$	(19.63)

Therefore, we have that

$\displaystyle[\mathbf{v}]_{\mathcal{B}}$	$\displaystyle=\begin{pmatrix}v_{1}\alpha_{1,1}+v_{2}\alpha_{2,1}+...+v_{\dim(V% )}\alpha_{\dim(V),1}\\ v_{2}\alpha_{1,2}+v_{2}\alpha_{2,2}+...+v_{\dim(V)}\alpha_{\dim(V),2}\\ ...\\ v_{\dim(V)}\alpha_{1,\dim(V)}+v_{\dim(V)}\alpha_{2,\dim(V)}+...+v_{\dim(V)}% \alpha_{\dim(V),\dim(V)}\end{pmatrix}$	(19.68)
	$\displaystyle=\begin{pmatrix}\alpha_{1,1}&\alpha_{2,1}&...&\alpha_{\dim(V),1}% \\ \alpha_{1,2}+\alpha_{2,2}&...&\alpha_{\dim(V),2}\\ ...\\ \alpha_{1,\dim(V)}&\alpha_{2,\dim(V)}&...&\alpha_{\dim(V),\dim(V)}\end{pmatrix% }\begin{pmatrix}v_{1}\\ v_{2}\\ ...\\ v_{\dim(V)}\end{pmatrix}$	(19.77)
	$\displaystyle=\begin{pmatrix}\alpha_{1,1}&\alpha_{2,1}&...&\alpha_{\dim(V),1}% \\ \alpha_{1,2}&\alpha_{2,2}&...&\alpha_{\dim(V),2}\\ ...\\ \alpha_{1,\dim(V)}&\alpha_{2,\dim(V)}&...&\alpha_{\dim(V),\dim(V)}\end{pmatrix% }[\mathbf{v}]_{\mathcal{C}}$	(19.82)

Alternatively - the change of basis matrix has as its $k$ th column the scalars needed to write the $k$ th element of the one basis as a linear combination of the others.

19.2.4 Subspaces

Definition 19.2.3

Let V be a vector space. The set W is a subspace of V if W is a vector space, and W is a subset of $V$ .

Technique 19.2.1

Showing that something is a subspace. Suppose we have a vector space V, and we want to prove that W is a subspace of V. The steps to do so are this

1.

Show that the zero vector is in the subspace in question.
2.

Show that $W\subseteq V$ using the standard technique for showing that something is a subset of something else (as in Section TODO: write).
3.

Then we must show that $W$ is closed under vector addition and scalar multiplication. The rest of the vector space axioms follow from the fact that $W\subseteq V$ and $V$ is a vector space.

This theorem is given as both an example of how to prove facts about vector spaces, but also because it is important in its own right.

Theorem 19.2.1

Let V be a vector space, and U and W be subspaces of V. Prove that $U\cup W$ is a subspace of $V$ if and only if $U\subseteq W$ or $W\subseteq U$ .

To prove this, first we will show the \sayif direction, and then the only if direction.

•

If. Without loss of generality, assume that $U\subseteq W$ , in which case $U\cup W=W$ , and this is a subspace of $V$ as $W$ is a subspace of $V$ . The proof for the other case follows by swapping $U$ and $W$ in the proof.
•
Only if. This direction requires a bit more of an intuition about what directions to explore. First we will assume that $U\cup W$ is a subspace, and then we will assume that the consequent is untrue (i.e. that $U\subseteq W\text{ or }W\subseteq U$ is not true), in which case there exist $u,w\in\textsf{V}$ such that

$u\in U\setminus W\text{ and }w\in W\setminus U.$ (19.83)

We can then ask (this is the core idea in the proof which is not immediately obvious – to me at least), about the status of $u+w$ . As $u,w\in U\cup W$ and by assumption $U\cup W$ is a subspace (and therefore by definition closed under vector addition) it must be that $u+w\in U\cup W$ . Then either $u+w\in U$ or $u+w\in W$ (by definition of the set union).
1. 1.
  
  If $u+w\in U$ , then also $u+w+(-u)=w$ which is a contradiction as by definition of $w$ (Equation LABEL:definition_of_u_and_w) $w\notin U$ .
2. 2.
  
  If $u+w\in W$ , a very similar thing is the case; also $u+w+(-w)=u\in W$ which is a contradiction as $u\notin W$ .
Therefore, by contradiction this direction of the theorem must be true.