# Day 12:

Linear Maps and their matrix representations

### Linear maps

Definition. Suppose $$V$$ and $$W$$ are vector spaces. A function $$L:V\to W$$ is called linear if for every $$x,y\in V$$ and $$a\in\mathbb{R}$$ the following two properties hold:

1. $$L(x+y) = L(x)+L(y)$$
2. $$L(ax)=aL(x)$$.

Example 1. Let  $A = \begin{bmatrix} 1 & 2 & 3\\ 0 & 1 & 0\end{bmatrix}.$ Define the function $$L:\mathbb{R}^{3}\to\mathbb{R}^{2}$$ by $$L(x) = Ax.$$ Using the properties of matrix multiplication we proved in Week 1, for any $$x,y\in\mathbb{R}^{3}$$ and $$a\in\mathbb{R}$$ we have $L(x+y) = A(x+y) = Ax+Ay = L(x) + L(y)$ and $L(ax) = A(ax) = aAx = aL(x).$ Thus, $$L$$ is a linear function.

### Linear maps

Example 2. Define the function $$D:\mathbb{P}_{2}\to\mathbb{P}_{1}$$ by $$Df(x) = f'(x)$$ for $$f(x)\in\mathbb{P}_{2}$$. For example,

$D(2x^2+x-1) = 4x+1.$

Using some properties of the derivative from calculus, if $$f(x),g(x)\in\mathbb{P}_{2}$$ and $$c\in\mathbb{R}$$, then

$D(f(x) + g(x)) = f'(x) + g'(x) = D(f(x)) + D(g(x))$

and

$D(cf(x)) = cf'(x) = cD(f(x)).$

Therefore, $$D$$ is linear.

Example 3.  The function $$F:\mathbb{R}\to\mathbb{R}$$ given by $$F(x) = 2x+1$$ is not linear, since

$F(1+1)=F(2) = 2(2)+1=5\neq 6 = F(1)+F(1).$

### Image and Kernel

Definition. Suppose $$V$$ and $$W$$ are vector spaces, and $$L:V\to W$$ is linear. The image of $$L$$ is the set

$\operatorname{im}(L) := \{Lv : v\in V\}.$

The kernel of $$L$$ is the set

$\operatorname{ker}(L):= \{v : L(v) = 0\}.$

The rank of $$L$$, denoted $$\operatorname{rank}(L)$$, is the dimension of the subspace $$\operatorname{im}(L)$$.

The nullity of $$L$$, denoted $$\operatorname{nullity}(L)$$, is the dimension of the subspace $$\operatorname{ker}(L)$$.

If $$A\in\mathbb{R}^{m\times n}$$, and we set $$L(x) = Ax$$ for $$x\in\mathbb{R}^{n}$$, then $$L:\mathbb{R}^{n}\to\mathbb{R}^{m}$$ is linear, moreover,

$\operatorname{im}(L) = C(A)\quad\text{and}\quad \operatorname{ker}(L) = N(A).$

### Surjective and Injective

Definition. Suppose $$V$$ and $$W$$ are vector spaces, and $$L:V\to W$$ is linear.

• The map $$L$$ is called injective (or one-to-one) if for every $$x,y\in V$$ such that $$x\neq y$$ it holds that $$L(x)\neq L(y)$$.
• The map $$L$$ is called surjective (or onto) if for every $$w\in W$$ there exists $$v\in V$$ such that $$L(v)=w$$.
• If $$L$$ is both injective and surjective, then we say that $$L$$ is bijective
• An injective map is called an injection. A surjective map is called a surjection. A bijective map is called a bijection.

Notes: A map $$L$$ is surjective if and only if $$\operatorname{im}(L)=W$$.

A map $$L$$ is injective if and only if $$L(x)=L(y)$$ implies $$x=y$$.

### Injections

Proposition. A linear map $$L:V\to W$$ is an injection if and only if $$\operatorname{ker}(L)=\{0\}$$.

Proof. Suppose $$L$$ is injective. Assume toward a contradiciton that there are two elements $$x,y\in\operatorname{ker}(L)$$ with $$x\neq y$$. This implies $$L(x)\neq L(y)$$, but $$L(x)=0=L(y)$$. This contradiction shows that $$\operatorname{ker}(L)$$ does not contain two distinct elements. Since $$0\in\operatorname{ker}(L)$$, we conclude that $$\operatorname{ker}(L)=\{0\}$$.

Next, suppose that $$\operatorname{ker}(L)=\{0\}$$. Let $$x,y\in V$$ such that $$L(x)=L(y)$$. By the linearity of $$L$$ we have

$0=L(x)-L(y) = L(x-y).$

This implies $$x-y\in\operatorname{ker}(L)$$, and hence $$x-y=0$$, or $$x=y$$. This shows that $$L$$ is injective. $$\Box$$

### Isomorphisms

Corollary. A linear map $$L:V\to W$$ is a bijection if and only if $$\operatorname{ker}(L)=\{0\}$$ and $$\operatorname{im}(L)=W$$.

Definition. A linear map $$L:V\to W$$ is called an isomorphism if $$\operatorname{ker}(L)=\{0\}$$ and $$\operatorname{im}(L)=W$$. If there exists an isomorphism from $$V$$ to $$W$$, then we say that $$V$$ and $$W$$ are isomorphic.

Example. Consider the map $$L:\mathbb{P}_{2}\to\mathbb{R}^{3}$$ given by

$L(ax^2+bx+c) = \begin{bmatrix}a\\ b\\ c\end{bmatrix}.$

We claim that this map is an isomorphism. First, note that if

$L(f(x)) = \begin{bmatrix} 0\\ 0\\ 0\end{bmatrix},$

then $$f(x) = 0$$. Thus $$L$$ is injective.

### Isomorphisms

Example continued. Next, let $$[a\ \ b\ \ c]^{\top}$$ be an arbitrary element of $$\mathbb{R}^{3}$$, then $$a,b,c\in\R$$ and we see that

$L(ax^2+bx+c) = [a\ \ b\ \ c]^{\top}.$

Thus, for any $$v\in\mathbb{R}^{3}$$ we see that there is some $$f(x)\in \mathbb{P}_{2}$$ such that $$L(f(x)) = v$$, that is, $$\operatorname{im}(L) = \mathbb{R}^{3}.$$ Hence, we see that $$\mathbb{P}_{2}$$ and $$\mathbb{R}^{3}$$ are isomorphic.

Theorem. For any $$n\in\N$$ the spaces $$\mathbb{P}_{n}$$ and $$\mathbb{R}^{n+1}$$ are isomorphic.

Proof. The map $$L:\mathbb{P}_{n}\to\mathbb{R}^{n+1}$$ given by

$L(a_{0} + a_{1}x+a_{2}x^2+\cdots+a_{n}x^{n}) = [a_0\ \ a_{1}\ \ a_{2}\ \ \cdots\ \ a_{n}]^{\top}$

is an isomorphism. (You should verify this) $$\Box$$

### Linear maps and bases

Suppose $$L:V\to W$$ is a linear map, and $$\{v_{i}\}_{i=1}^{n}$$ is a basis for $$V$$.

If you only know the outputs $$L(v_{1}),L(v_{2}),\ldots,L(v_{n})$$, then you know any output of the function $$L$$.

Indeed, given $$v\in V$$, there exist unique scalars $$a_{1},a_{2},\ldots,a_{n}$$ such that

$v = a_{1}v_{1} + a_{2}v_{2} + \cdots + a_{n}v_{n}.$

By linearity we have

$L(v) = L(a_{1}v_{1} + a_{2}v_{2} + \cdots + a_{n}v_{n}) = a_{1}L(v_{1}) + a_{2}L(v_{2}) + \cdots + a_{n}L(v_{n}).$

We know all the vectors $$L(v_{1}),L(v_{2}),\ldots,L(v_{n}),$$ so we can compute $$L(v)$$.

### Linear maps and bases

Suppose $$L:V\to W$$ is a linear map, $$\{v_{i}\}_{i=1}^{n}$$ is a basis for $$V$$, and $$\{w_{i}\}_{i=1}^{m}$$ is a basis for $$W$$.

To know $$L$$ we need to know the vectors $$L(v_{1}),L(v_{2}),\ldots,L(v_{n})$$. For each $$j\in\{1,2,\ldots,n\}$$ there are unique scalars $$a_{1j},a_{2j},a_{3j},\ldots,a_{mj}$$ such that

$L(v_{j}) = \sum_{i=1}^{m}a_{ij}w_{i}.$

Therefore, in order to know the map $$L$$, we only need to know the numbers $$\{a_{ij}\}_{i=1,j=1}^{m,n}$$. It's easier to organize this into a grid:

$\begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{m1} & a_{m2} & \cdots & a_{mn}\end{bmatrix}$

### Linear maps and bases

The matrix

$\begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{m1} & a_{m2} & \cdots & a_{mn}\end{bmatrix}$

is called the matrix representation of $$L$$ with respect to the bases $$\{v_{i}\}_{i=1}^{n}$$ and $$\{w_{i}\}_{i=1}^{m}$$.

### Matrix representation

Example. Let $$D:\mathbb{P}_{2}\to\mathbb{P}_{1}$$ be given by $D(f(x)) = f'(x).$

Fix the bases $$S = \{1,x,x^2\}$$ for $$\mathbb{P}_{2}$$ and $$T = \{1,x\}$$ for $$\mathbb{P}_{1}$$.

We compute $$D(v)$$ for each $$v\in S$$, and write the output as a linear combination of the vectors in $$T$$:

$D(1) = 0 = 0\cdot 1 + 0\cdot x$

$D(x) = 1 = 1\cdot 1 + 0\cdot x$

$D(x^{2}) = 2x = 0\cdot 1 + 2\cdot x$

Then, the matrix representation of $$D$$ with respect to $$S$$ and $$T$$ is

$\begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 2\end{bmatrix}$

By John Jasper

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