19.3 Linear transformations

Proof: there are two directions to show.

1. 1.
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   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.

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 $\mathcal{B}=\{\beta_{i}\}_{1\leqq i\leqq n}$.

• •
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  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

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

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

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

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

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 $T(\beta_{j})$.