A Linear Map (a.k.a. Linear Transformation) is a function which maps elements between two vector space that follows linear properties.

## constituents

- vector spaces \(V\) and \(W\) (they don’t have to be subspaces)
- A function \(T: V \to W\) (when we put something in, it only goes to one place)

## requirements

\(T\) is considered a Linear Map if it follows… (properties of “linearity”)

### additivity

\begin{equation} T(u+v) = Tu+Tv,\ \forall u,v \in V \end{equation}

### homogeneity

\begin{equation} T(\lambda v) = \lambda (Tv),\ \forall \lambda \in \mathbb{F}, v \in V \end{equation}

## additional information

### note on notation

The “application” of a Linear Map \(T\) on a vector \(V\) is written as:

\begin{equation} \begin{cases} Tv \\ T(v) \end{cases} \end{equation}

both are acceptable.

### \(\mathcal{L}(V,W)\)

The set of all Linear Maps from \(V\) to \(W\) is denoted as \(\mathcal{L}(V,W)\).

### some fun linear maps

#### zero

There is, of course, the Linear Map that maps everything to the \(0\) — as the zero exists in all vector spaces.

That is:

\begin{equation} 0 \in \mathcal{L}(V,W) \implies 0v = 0, v\in V \end{equation}

additivity

Let \(v_1+v_2= v \in V\).

\begin{equation} 0(v_1+v_2) = 0(v) = 0 = 0+0 = 0v_1+0v_2 \end{equation}

Let \(\lambda v = u \in V\).

\begin{equation} 0(\lambda v) = 0(u) = 0 = \lambda 0 = \lambda 0v \end{equation}

#### identity

Another classic. \(I\), the identity map, is denoted as (for some \(v \in V\) and \(I \in \mathcal{L}(V,V)\)):

\begin{equation} Iv = v \end{equation}

i.e. it does nothing

additivity

Let \(v_1,v_2 \in V\):

\begin{equation} I(v_1+v_2) = v_1+v_2 = Iv1+Iv2 \end{equation}

\begin{equation} I(\lambda v) = \lambda v = \lambda Iv \end{equation}

#### any map from \(\mathbb{F}^{n}\) to \(\mathbb{F}^{m}\)

turns out any map that follows a specific pattern of polynomials between two vector spaces are Linear Maps.

Define some two vector spaces \(\mathbb{F}^{n}\) and \(\mathbb{F}^{m}\), some set of scalars \(a_{jk} \in \mathbb{F}: j=1, \dots m; k=1, \dots n\).

We construct \(T \in \mathcal{L}(\mathbb{F}^{n}, \mathbb{F}^{m})\) by: \(T(x_1, \dots x_{n}) = a_{11} x_1+ \cdots + a_{1n} x_{n}, \dots, a_{m1} x_1 + \cdots + a_{mn} x_{n}\) (i.e. a combination of linear combinations).

additivity

Let \(x,y \in \mathbb{F}^{n}\), with \(x_{j}\) being each coordinate of \(x\) and the same goes for \(y\).

\begin{align} T((x_1, \dots x_{n}) + (y_1, \dots y_{n}))) &= T(x_1+y_1 \dots x_{n}+y_{n}) \\ &= a_{11}(x_1+y_1) + \cdots, \dots, \cdots + a_{mn} (x_{n} + y_{n}) \\ &= (a_{11}x_1 + a_{11}y_{1}) + \cdots, \dots, \cdots + (a_{mn} x_{n} + a_{mn} y_{n}) \\ &= (a_{11}x_1 + \cdots) + (a_{11}y_{1}+ \cdots ), \dots, (\cdots + a_{mn}x_n) + (\cdots + a_{mn}y_{n}) \\ &= ((a_{11}x_1 + \cdots), \dots, (\cdots + a_{mn}x_n)) = ((a_{11}y_{1}+ \cdots ), \dots,(\cdots + a_{mn}y_{n})) \\ &= T (x_1, \dots, x_{n}) + T (y_1, \dots, x_{n}) \end{align}

Proof left to the reader. Pretty much just expand and more algebra.

### matricies to encode Linear Map

we can use matricies to represent Linear Maps. See matrix of Linear Map

### “basis of domain”

This result tells us that we can find a Linear Map for wherever we want to take the basis of a vector space, and that a Linear Map’s behavior on basis uniquely determines that Linear Map.

See basis of domain.

### addition and scalar multiplication on \(\mathcal{L}(V,W)\)

Suppose \(S,T \in \mathcal{L}(V,W); \lambda \in \mathbb{F}\).

“Sum” and “Product” are defined in the way that one would expect:

\begin{equation} (S+T)(v) = Sv+Tv \end{equation}

and

\begin{equation} (\lambda T)(v) = \lambda (Tv) \end{equation}

for all \(v \in V\).

These two operations make \(\mathcal{L}(V,W)\) a vector space (\(1Tv = Tv\), \(0+Tv=Tv\), \(Tv + (-1)Tv = 0\), associativity, commutativity, distributive inherits from \(V\).)

### linear maps take \(0\) to \(0\)

We desire that \(T(0) = 0\) for any linear map \(T\)

Proof:

\begin{equation} T(0) = T(0+0) \end{equation}

Then, by additivity:

\begin{equation} T(0) = T (0 + 0) = T (0) + T (0) \end{equation}

Given \(\mathcal{L}(V,W)\) is a vector space for any \(V,W\), \(\exists -T(0)\) such that \(T(0)+(-T(0)) = 0\). Applying that here:

\begin{equation} T(0) = T(0)+T(0) \implies T(0) -T(0) = T(0)+T(0)-T(0) \implies 0 = T(0) \end{equation}

### Product of Linear Maps

### “sizes” of maps

#### map to smaller space is not injective

Suppose \(V,W\) are finite-dimensional vector spaces, and \(\dim V > \dim W\). Then, all \(T \in \mathcal{L}(V,W)\) are **not** injective.

We first have that:

\begin{align} &\dim V = \dim null\ T + \dim range\ T \\ \Rightarrow\ & \dim null\ T = \dim V - \dim range\ T \end{align}

recall at this point that \(\dim range\ T \leq \dim W\) (the range is a subspace of the codomain.) Therefore, subtracting a bigger value means that the value will be smaller. So we have that:

\begin{align} & \dim null\ T = \dim V - \dim range\ T \\ \Rightarrow\ & \dim null\ T \geq \dim V - \dim W \end{align}

Now, recall that \(\dim V > \dim W\). Therefore, \(\dim V - \dim W\) is strictly bigger than \(0\). So:

\begin{align} \dim null\ T &\geq \dim V - \dim W \\ &> 0 \end{align}

And so, the dimension of the null space of \(T\) is not \(0\). Therefore, the null space of \(T\) can’t have been \(\{0\}\) because that does have dimension \(0\). This makes the map not injective because injectivity implies that null space is \(\{0\}\)

#### map to bigger space is not surjective

Its basically the same thing as the one above. Suppose \(V,W\) are finite-dimensional vector spaces, and \(\dim V < \dim W\). Then, all \(T \in \mathcal{L}(V,W)\) are **not** injective.

We first have that:

\begin{align} &\dim V = \dim null\ T + \dim range\ T \\ \Rightarrow\ & \dim range\ T = \dim V - \dim null\ T \end{align}

Because the dimension of \(null\ T\) is larger than \(0\) (or, for that matter, the dimension of anything), \(\dim V - \dim\ null\ T \leq \dim\ V\). Hence:

\begin{align} & \dim range\ T = \dim V - \dim null\ T \\ \Rightarrow\ & \dim range\ T \leq \dim V \end{align}

Now, recall that \(\dim V < \dim W\).

\begin{align} & \dim range\ T = \dim V - \dim null\ T \\ \Rightarrow\ & \dim range\ T \leq \dim V < \dim W \end{align}

Given the range of \(T\) is smaller than the codomain of \(T\), they cannot be equal spaces. So, \(T\) is not surjective.