Lecture 13 - Linear Transformations

Studying vectors on their own is boring. Now, we'll look at mappings from vectors to other vectors.

For example, consider $V = P (R)$ , the space of polynomials with real coefficients. This is a countably-infinitely-dimensional vector space.

Now let $D$ denote the calculus derivative map. So then $D : V \to V$ such that it's defined by:

D (p \in V) = p^{'}

and if you remember power rule, we know how to take the derivative of a polynomial. In fact, we know that taking this derivative would be closed under derivation, as we get a polynomial as an output.

$D$ actually has some really nice properties:

$D (p + q) = D p + D q$ . This comes from calculus where $(p + q)^{'} = p^{'} + q^{'}$ .
$D (α p) = α D p$ . This comes from calculus where $(α p)^{'} = α p^{'}$ .
Because these 2 properties are satisfied, then we say $D$ is a linear map or a linear transformation. Sometimes you'll hear that $D$ is "linear", and that's that this means.

These two properties come up so often that we'll generalize it:

Linear Transformation

Given a vector space $V$ and a vector space $W$ , where $V, W$ are over the same field $F$ , then the function $T : V \to W$ is a linear transformation iff:

$T (v_{1} + v_{2}) = T v_{1} + T v_{2}$ for all $v_{1}, v_{2} \in V$
$T (α v) = α T v$ for all $α \in F, v \in V$

Note that there are two different $+$ signs in (1). The $v_{1} + v_{2}$ adds vectors in $V$ , while $T v_{1} + T v_{2}$ are adding vectors in $W$ .

Some notation: sometimes you want to have the collection of linear maps from $V$ to $W$ . We denote it as $L (V, W)$ to denote this collection of linear maps from $V$ to $W$ . Furthermore, if $V$ is $W$ , then instead of saying $L (V, V)$ we can just say $L (V)$ .

So then, as an example, $D \in L (P (F))$ .

But, okay this is mind blowing, $L (V, W)$ itself is a vector space. That's because:

The zero function $Z (v) = \vec{0}$ is in our space.
Adding two functions $A, B \in L (V, W)$ gives a function that's also a linear map.
There's an additive inverse $- A$ for every $A$
...

Okay let's see this in more detail. If we had some function $f$ where $f (v) = w$ then the additive inverse $- f$ would be the function where $(- f) (v) = - w$ . Or, given a function $T \in L (V, W)$ , define $S$ by $S (v) = - T (v)$ . Since $S + T = Z$ then $T$ is the additive inverse of $S$ and vice versa.

Properties of Linear Maps

Here are some properties of linear maps:

Properties of Linear Maps

For the following, we say $T \in L (V, W)$ , or $T \in L (V, W)$ .

$T ({\vec{0}}_{V}) = {\vec{0}}_{W}$ . This is because $T (v - v) = T v - T v = {\vec{0}}_{W}$ , or $T (0 v) = 0 T (v) = {\vec{0}}_{W}$ . Notice that the zero vectors may be different, as they may come from different vector spaces.
$T (\sum c_{i} {\vec{v}}_{i}) = \sum c_{i} T ({\vec{v}}_{i})$ .
$T (v_{1} - v_{2}) = T (v_{1}) - T (v_{2})$ .

Let's look at an example. Consider $R^{2}$ , the transformation $C$ rotates an input vector CCW by $α$ radians. We can show that this is linear transformation, via a picture:

You could check property 2 (scalar multiplication) too and get that that holds too.

Let's look at another example. Let's work in $C (R)$ , the space of continuous functions with domain $R$ . We'll define a function $T : C (R) \to R$ , defined by:

T (f) = \int_{a}^{b} f (x) d x

where $a, b$ are fixed, real numbers. We claim that $T$ is linear:

$T (f + g) = \int_{a}^{b} (f + g) (x) d x = \int_{a}^{b} f (x) d x + \int_{a}^{b} g (x) d x = T (f) + T (g)$ .
$T (α f) = \int_{a}^{b} (α f) (x) d x = α \int_{a}^{b} f (x) d x = α T (f)$ .

So if we look at $L (C (R), R)$ , then $T \in L (C (R), R)$ . But notice that here $a, b$ were arbitrary. If we have another one where we have $b, c$ defined in $T^{'}$ then it's still a linear transformation. But if we consider $T + T^{'}$ then that itself is still a linear transformation! You can see how meta this can get...

LT's Defined by Their Bases

Linear Transformations Defined on the Actions of their Basis

Given a vector space $V$ with basis $v_{1}, . . ., v_{n}$ , then given any vectors $w_{1}, . . ., w_{n}$ , in a vector space $W$ (all over the same field $F$ ), there is a unique transformation $T : V \to W$ satisfying:

T (v_{i}) = w_{i}

for all $i$ .

We'll do part of the proof today:

Proof
This is an existence/uniqueness proof. We'll have to define a linear transformation. Suppose $T$ were a LT with the property that $T (v_{i}) = w_{i}$ for all $i$ . Let $v \in V$ be an arbitrary vector. Then since $v_{1}, . . ., v_{n}$ is a basis, there's a linear combination of the basis elements that construct $v$ :

a_{1} v_{1} + \dots + a_{n} v_{n} = v

where each $a_{i}$ exists and is unique. Then:

T (v) = T (a_{1} v_{1} + \dots + a_{n} v_{n}) = a_{1} T (v_{1}) + \dots + a_{n} T (v_{n})

but since $T (v_{i}) = w_{i}$ :

v = a_{1} w_{1} + \dots + a_{n} w_{n}

so if there were two $T_{1}, T_{2}$ , doing this process yeilds that they are the same.

But is there one? We can just define $T : V \to W$ as:

T (a_{1} v_{1} + \dots + a_{n} v_{n}) = a_{1} w_{1} + \dots + a_{n} w_{n}

and all we would have to show is that $T$ is linear and $T (v_{i}) = w_{i}$ for all $i$ . The second just comes from the definition, but the linear part requires a bit more work.

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