# 19.3 Linear transformations

## 19.3.1 Introduction to linear transformations

###### Definition 19.3.1

Let $\textsf{V},\textsf{W}$ be vector spaces over a field $\mathbb{K}$. We say a function $T:\textsf{V}\to\textsf{W}$ is linear if it satisfies these two properties

1. 1.

For every $x,y\in\textsf{V}$,

 $\displaystyle T(x+y)=T(x)+T(y)$ (19.84) $\displaystyle T(\alpha x)=\alpha T(x)$ (19.85)
###### Theorem 19.3.1

Let $\textsf{V},\textsf{W}$ be vector spaces over a field $\mathbb{K}$ and let $x,y\in\textsf{V}$ and $\alpha\in\mathbb{K}$. The map/function/ whatever you want to call it $T:\textsf{V}\to\textsf{W}$ is linear if and only if

$T(\alpha x+y)=\alpha T(x)+T(y)$ (19.86)

Proof: there are two directions to show.

1. 1.

Only if. We assume that $T$ is a linear transformation. Therefore, $T$ satisfies 19.84, so we can write

 $\displaystyle T(\alpha x+y)$ $\displaystyle=T(\alpha x)+T(y)$ (19.87)

As $T$ is linear it also satisifies 19.85, so by this property,

 $\displaystyle T(\alpha x+y)$ $\displaystyle=\alpha T(x)+T(y)$ (19.89)
2. 2.

If. We assume that 19.86 holds, and therefore (this is just a restatement of the equation from the theorem)

 $\displaystyle T(\alpha x+y)=T(\alpha x)+T(y)$ (19.90)

We then set $\alpha=1$, so it follows that

$T(x+y)=T(x)+T(y)$ (19.91)

What remains to show is that for all $\alpha$ and $x$ we have

$T(\alpha x)=\alpha T(x)$ (19.92)

We obtain this by fixing $y=0$ from which the result for all $\alpha$ and $\mathbb{K}$ follows.

## 19.3.2 The matrix representation of a linear transformation

The best way to understand this is to do a lot of examples, with specific linear transformations and vector spaces. It’s easy to get lost, sinking, \saynot waving but drowning in the steaming soup of generality. As they don’t say, a little reification 77 7 Meaning turning something abstract into something concrete. every day keeps the doctor away.

Let’s assume that we have a linear transformation $T:V\to W$, and we would like to find its matrix representation. It’s really easy to get confused here, but don’t lose sight of the goal. We need some information about $V$ and $W$, specifically

• A basis for $V$, denoted as $\mathcal{B}=\{\beta_{i}\}_{1\leqq i\leqq n}$.

• A basis for $W$, denoted as $\mathcal{C}=\{\gamma_{i}\}_{1\leqq i\leqq m}$.

We then pick an arbitrary vector, $\mathbf{v}\in V$, and finds its representation as a linear combination of $\beta$, that is we find

$\mathbf{v}=\sum_{1\leqq i\leqq n}v_{i}b_{i}$ (19.93)

But we’re not after $\mathbf{v}$, we’re after $T(\mathbf{v})$! Therefore, we apply $T$ to both sides, giving

 $\displaystyle T(\mathbf{v})$ $\displaystyle=T\left(\sum_{1\leqq j\leqq n}v_{j}\beta_{j}\right)$ (19.94)

We now use liberally the fact that $T$ is linear.

 $\displaystyle T(\mathbf{v})$ $\displaystyle=\sum_{1\leqq j\leqq n}v_{j}T(\beta_{j})$ (19.96)

We’re not dealing with a concrete linear transformation, so \sayall we can say is that for each $i$, $T(\beta_{j})$ will give us a vector in $W$ and that we can certainly write this as a linear combination of $\mathcal{C}$, as it is a basis for $W$. Every $T(\beta_{j})$ is a linear combination of the $m$ vectors in $\mathcal{C}$, i.e. $T(\beta_{j})=\sum_{1\leqq i\leqq m}\left(a_{i,j}\right)\gamma_{i}$. Substituting this in, we get

 $\displaystyle T(\mathbf{v})$ $\displaystyle=\sum_{1\leqq j\leqq n}v_{j}\left(\sum_{1\leqq i\leqq m}a_{i,j}% \gamma_{i}\right)$ (19.97) $\displaystyle=\sum_{1\leqq i\leqq m}\gamma_{i}\left(\sum_{1\leqq j\leqq n}a_{i% ,j}v_{j}\right)$ (19.98) $\displaystyle=\sum_{1\leqq i\leqq m}\left(\sum_{1\leqq j\leqq n}a_{i,j}v_{j}% \right)\gamma_{i}$ (19.99)

Now, from the definition of a co-ordinate vector, as $\gamma_{1},...,\gamma_{m}$ are the basis vectors for $\mathcal{C}$, the representation of $T(v)$ as a co-ordinate vector in this basis is just

 $\displaystyle[T(v)]_{\mathcal{C}}$ $\displaystyle=\begin{pmatrix}\sum_{1\leqq j\leqq n}a_{1,j}v_{j}\\ \sum_{1\leqq j\leqq n}a_{2,j}v_{j}\\ ...\\ \sum_{1\leqq j\leqq n}a_{m,j}v_{j}\\ \end{pmatrix}$ (19.104) $\displaystyle=\underbrace{\begin{pmatrix}a_{1,1}&a_{1,2}&...&a_{1,n}\\ a_{2,1}&a_{2,2}&...&a_{2,n}\\ ...&...&...&...\\ a_{m,1}&a_{1,2}&...&a_{1,n}\end{pmatrix}}_{\text{Let this be \mathbf{A}.}}% \begin{pmatrix}v_{1}\\ v_{2}\\ ...\\ v_{3}\end{pmatrix}$ (19.113)

Which is exactly what we wanted to find. Specifically, the $j$th column of the matrix $\mathbf{A}$ (as defined in Equation 19.113) is the co-ordinate vector (in the ordered base $\mathcal{C}$) of the result of $T(\beta_{j})$.