# 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. 1.

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

2. 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. 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. 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. 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. 1.

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

2. 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. 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.