Let $F:V\longrightarrow W$ be a linear map. The image of $F$ is the set
$$\mathrm{Im}F=\{w\in W: F(v)=w\ \mbox{for some}\ v\in V\}.$$

Proposition. The image of $F$ is a subspace of $W$.

Proof. The proof is straightforward. It is left for an exercise.

The preimage of the identity element $O$ under the linear map $F$ i.e. the set of elements $v\in V$ such that $F(v)=O$ is called the kernel of $F$ and is denoted by $\ker F$.

Proposition. The kernel of $F$ is a subspace of $V$.

Proof. It is straightforward and left for an exercise.

Example. Let $L: \mathbb{R}^3\longrightarrow\mathbb{R}$ be the map defined by
$$L(x,y,z)=3x-2y+z.$$
If we write $A=(3,-2,1)$, then $L(X)$ may be written as
$$L(X)=X\cdot A.$$
So, the kernel of $L$ is the set of all $X$ that are perpendicular to $A$.

Example. Let $A$ be an $m\times n$ matrix and let $L_A:\mathbb{R}^n\longrightarrow\mathbb{R}^m$ be the linear map defined by
$$L_A(X)=AX.$$
The kernel of $L_A$ is the subspace of solutions $X$ of the system of linear equations
$$AX=O.$$

Example. Let $\mathcal{F}$ be the vector space of smooth functions. Let $a_1,\cdots,a_m$ be numbers and let
$$L=a_m\frac{d^m}{dx^m}+a_{m-1}\frac{d^{m-1}}{dx^{m-1}}+\cdots+a_1.$$
Then $L:\mathcal{F}\longrightarrow\mathcal{F}$ is a linear map. $\ker L$ is the space of solutions of the homogeneous linear differential equation
$$a_m\frac{d^mf}{dx^m}+a_{m-1}\frac{d^{m-1}f}{dx^{m-1}}+\cdots+a_1f=0.$$
If there exists one solution $h_0$ for the non-homogeneous linear differential equation $L(h)=g$, then any solution $h$ may be written as $h=f+h_0$ where $f$ is a solution of the homogeneous equation $L(f)=0$. The proof is left as an exercise.

Theorem. Let $F: V\longrightarrow W$ be a linear map such that $\ker F=\{O\}$. If $v_1,\cdots,v_n$ are linearly independent elements of $V$, then $F(v_1),\cdots,F(v_n)$ are linearly independent elements of $W$.

Proof. The proof is straightforward and is left for an exercise.

Theorem. Let $F: V\longrightarrow W$ be a linear map. $F$ is one-to-one if and only if $\ker F=\{O\}$.

Proof. Suppose that $F$ is one-to-one. Let $v\in\ker F$. Then $F(v)=O=F(O)$. Since $F$ is one-to-one, $v=O$. So, $\ker F=\{O\}$. Suppose that $\ker F=\{O\}$. Let $F(v_1)=F(v_2)$. Then $F(v_1-v_2)=O$ and so $v_1-v_2\in\ker F=\{O\}$ i.e. $v_1=v_2$. Hence, $F$ is one-to-one.

Given a linear map $L: V\longrightarrow W$, there is a relationship between the dimensions of $V$, $\ker L$, and $\mathrm{Im}L$, namely
$$\dim V=\dim\ker L+\dim\mathrm{Im}L.$$
We will not prove it here but those interested may find proof in [1].

Example. Consider the linear map $L:\mathbb{R}^3\longrightarrow\mathbb{R}$ given by
$$L(x,y,z)=3x-2y+z.$$
The image is not $\{O\}$, so it is $\mathbb{R}$. Therefore the dimension of $\ker L$, the space of all solutions of $3x-2y+z=0$ is 2. $3x-2y+z=0$ is indeed equation of a plane through the origin and we know that the dimension of a plane as a vector space is 2.

References:

[1] Serge Lang, Introduction to Linear Algebra, Second Edition, Undergraduate Texts in Mathematics, Springer, 1986

Let $V, W$ be two vector spaces. A map $L:V\longrightarrow W$ is called a linear map if it satisfies the following properties: for any elements $u,v\in V$ and any scalar $c$,

LM 1. $L(u+v)=L(u)+L(v)$.

LM 2. $L(cu)=cL(u)$.

That is, linear maps are maps that preserve addition and scalar multiplication.

Proposition. A map $L: V\longrightarrow W$ is linear if and only if for any elements $u,v\in V$ and scalars $a,b$,
$$L(au+bv)=aL(u)+bL(v).$$

Proof. It is straightforward and left as an exercise.

Example. Let $A$ be an $m\times n$ matrix. Define
$$L_A:\mathbb{R}^n\longrightarrow\mathbb{R}^m$$
by
$$L_A(X)=A\cdot X.$$ Then $L_A$ is linear.

Example. Let $A=(a_1,\cdots,a_n)$ be a fixed vector in $\mathbb{R}^n$. Define $L_A:\mathbb{R}^n\longrightarrow\mathbb{R}$ by
$$L_A(X)=A\cdot X.$$
Then $L_A$ is a linear map. The dot product $A\cdot X$ can be viewed as a matrix multiplication if we view $A$ as a row vector and $X$ as a column vector. So this example is a spacial case of the previous example.

Example. Let $\mathcal{F}$ be the set of all smooth functions. Then the derivative $D:\mathcal{F}\longrightarrow\mathcal{F}$ is a linear map.

Example. Define $\wp: \mathbb{R}^3\longrightarrow\mathbb{R}^2$ by $\wp(x,y,z)=(x,y)$, i.e. $\wp$ is a projection. It is a linear map.

Proposition. Let $L: V\longrightarrow W$ be a linear map. Then $L(O)=O$.

Proof. Let $v\in V$. Then
$$L(O)=L(v-v)=L(v-v)=L(v)-L(v)=O.$$

Example. Let $L:\mathbb{R}^2\longrightarrow\mathbb{R}^2$ be a linear map. Suppose that
$$L(1,1)=(1,4)\ \rm{and}\ L(2,-1)=(-2,3).$$
Find $L(3,-1)$.

Solution. $(3,-1)$ is written as a linear combination of $(1,1)$ and $(2,-1)$ as
$$(3,-1)=\frac{1}{3}(1,1)+\frac{4}{3}(-2,3).$$
Hence,
$$L(3,1)=\frac{1}{3}L(1,1)+\frac{4}{3}L(-2,3)=\frac{1}{3}(1,4)+\frac{4}{3}(-2,3)=\left(-\frac{7}{3},\frac{16}{3}\right).$$

The coordinates of a linear map

Consider a map $F: V\longrightarrow\mathbb{R}^n$. For any $v\in V$, $F(v)\in\mathbb{R}^n$ so $F(v)$ may be written as
$$F(v)=(F_1(v),F_2(v),\cdots,F_n(v))$$
where each $F_i$ is a function $F_i:V\longrightarrow\mathbb{R}$ called the $i$-th coordinate function.

Proposition. A map $F_i: V\longrightarrow\mathbb{R}^n$ is linear if and only if each coordinate function $F_i$ is linear.

Proof. Straightforward. Left as an exercise.

Example. Let $F:\mathbb{R}^2\longrightarrow\mathbb{R}^3$ be the map
$$F(x,y)=(2x-y,3x+4y,x-5y).$$
Then
$$F_1(x,y)=2x-y,\ F_2(x,y)=3x+4y,\\ F_3(x,y)=x-5y.$$
These coordinate functions can be written as
$$F_1(x,y)=\begin{pmatrix}
2 & -1
\end{pmatrix}\begin{pmatrix}
x\\y
\end{pmatrix},\ F_2(x,y)=\begin{pmatrix}
3 & 4
\end{pmatrix}\begin{pmatrix}
x\\y
\end{pmatrix},\ F_3(x,y)=\begin{pmatrix}
1 & -5
\end{pmatrix}\begin{pmatrix}
x\\y
\end{pmatrix}.$$
Hence, each $F_i$ is linear, $i=1,2,3$ and therefore $F$ is linear by the Proposition. In fact, $F$ may be written as $L_A:\mathbb{R}^2\longrightarrow\mathbb{R}^3$ where
$$A=\begin{pmatrix}
2 & -1\\
3 & 4\\
1 & -5
\end{pmatrix}.$$

Consider an $m\times n$ matrix

$$A=\begin{pmatrix}a_{11} & \cdots & a_{1n}\\\vdots & & \vdots\\a_{m1} & \cdots & a_{mn}\end{pmatrix}.$$

The columns of $A$ generate a vector space, which is a subspace of $\mathbb{R}^m$, called the column space of $A$. The dimension of the subspace is called the column rank of $A$. Similarly the rows of $A$ generate a subspace of $\mathbb{R}^n$, called the row space of $A$ and the dimension of this subspace is called the row rank of $A$. It turns out that the column rank and the row rank must be equal. So, we simply call the column rank or the row rank of $A$, the rank of $A$.

There are a couple important theorems regarding the rank of a matrix. They are introduced without proofs.

Theorem. Row and column operations do not change the row rank of a matrix, nor do they change the column rank.

Remark. Row and column operations only change basis of row space or column space.

Theorem. Let $A$ be a matrix of rank $r$. By a succession of row and column operations, the matrix can be transformed to the matrix having components equal to $1$ on the diagonal of the first $r$ rows and columns, and $0$ everywhere else.

$$\begin{pmatrix}1 & 0 & \cdots & 0 & 0 &\cdots &0\\0 & 1 & \cdots & 0 & 0 & \cdots &0\\\vdots & &\ddots &\vdots&\vdots& &\vdots\\0 & 0 &\cdots & 1& 0 &\cdots &0\\0 & 0 &\cdots & 0& 0 &\cdots &0\\\vdots & & &\vdots&\vdots&\ddots&\vdots\\0 & 0 &\cdots & 0& 0 &\cdots &0\end{pmatrix}$$

Example. Find the rank of the matrix $\begin{pmatrix}2 & 1 & 1\\0 & 1 & -1\end{pmatrix}$.

Solution. There are only two rows, so the rank will be at most 2. On the other hand, the column vectors $\begin{pmatrix}2\\0\end{pmatrix}$ and $\begin{pmatrix}1\\1\end{pmatrix}$ are linear independent. Therefore, the rank is 2.

Example. Find the rank of the matrix

$$\begin{pmatrix}1 & 2 & -3\\2 & 1 & 0\\-2 & -1 & 3\\-1 & 4 & -2\end{pmatrix}.$$

Solution. Since there are three columns, the rank will be at most 3. Subtract 2 times column 1 from column 2; add 3 times column 1 to column 3. The resulting matrix is

$$\begin{pmatrix}1 & 0 & 0\\2 & -3 & 6\\-2 & 3 & -3\\-1 & 6 & -5\end{pmatrix}.$$

Add 2 times column 2 to column 3. The resulting matrix is

$$\begin{pmatrix}1 & 0 & 0\\2 & -3 & 0\\-2 & 3 & 3\\-1 & 6 & 7\end{pmatrix}.$$

This matrix is in column echelon form and one can easily see that the first three rwo vectors are linearly independent. Therefore, the rank is 3.

Let $V$ be a vector space. $v_1,\cdots,v_n\in V$ are said to be linearly dependent if there exist numbers $a_1,\cdots,a_n$ not all equal to $0$ such that
$$a_1v_1+\cdots+a_n=O.$$
If there do not exist such numbers, then we say $v_1,\cdots,v_n$ are linearly independent. That is, $v_1,\cdots,v_n$ are linearly independent if whenever $a_1v_1+\cdots+a_nv_n=O$, $a_1=\cdots=a_n=0$.

Example. In $\mathbb{R}^n$, the standard unit vectors $E_1,\cdots,E_n$ are linearly independent.

Example. The vectors $(1,1)$ and $(-3,2)$ are linearly indepdent in $\mathbb{R}^2$.

A set of vectors $\{v_1,\cdots,v_n\}\subset V$ is said to be a basis of $V$ if $v_1,\cdots,v_n$ generate $V$ and that they are linearly independent.

Example. The vectors $E_1,\cdots,E_n$ form a basis of $\mathbb{R}^n$.

Example. The vectors $(1,1)$ and $(-1,2)$ form a basis of $\mathbb{R}^2$.

In general for $\mathbb{R}$, the following theorem holds.

Theorem. Let $(a,b)$ and $(c,d)$ be two vectors in $\mathbb{R}^2$.

(i) They are linearly depedendent if and only if $ad-bc=0$.

(ii) If they are  indepdendent, they form a basis of $\mathbb{R}^2$.

Proof. Exercise

Let $V$ be a vector space and let $\{v_1,\cdots,v_n\}$ be a basis of $V$. If $v\in V$ is written as a linear combination
$$v=x_1v_1+\cdots+x_nv_n,$$
$(x_1,\cdots,x_n)$ is called the coordinates of $v$ with respect to the basis $\{v_1,\cdots,v_n\}$. For each $i=1,\cdots,n$, $x_i$ is called the i-th coordinate. The following theorem says that there can be only one set of coordinates for a given vector.

Theorem. Let $V$ be a vector space. Let $v_1,\cdots,v_n$ be linearly independent elements of $V$. If $x_1,\cdots,x_n$ and $y_1,\cdots,y_n$ are numbers such that
$$x_1v_1+\cdots+x_nv_n=y_1v_1+\cdots+y_nv_n,$$
then $x_i=y_i$ for all $i=1,\cdots,n$.

Proof. It is strightforward from the definition of linearly independent vectors.

Example. Find the coordinates of $(1,0)$ with respect to the two vectors $(1,1)$ and $(-1,2)$.

Example. The two functions $e^t$ and $e^{2t}$ are linearly independent.

Proof. Exercise.

Theorem. Let $v,w$ be two vectors of a vector space $V$. They are linearly dependent if and only if one of them is a scalr multiple of the other, i.e there is a number $c$ such that $v=cw$ or $w=cv$.

Proof. Exercise.

If one basis of a vector space $V$ has $n$ elements and another basis has $m$ elements, then $n=m$. The number of elements in any basis of a vector space $V$ is called the dimension of $V$ and is denoted by $\dim V$.

Let $V$ be a vector space and let $v_1,\cdots, v_n\in V$. $V$ is said to be generated by $v_1,\cdots,v_n$ if given an element $v\in V$, there exist numbers $x_1,\cdots, x_n$ such that
$$v=x_1v_1+\cdots+x_nv_n.$$
The expression $x_1v_1+\cdots+x_nv_n$ is called a linear combination of $v_1,\cdots,v_n$. The numbers $x_1,\cdots,x_n$ are called the coefficients of the linear combination.

Example. Let $E_1,\cdots,E_n$ be the standard unit vectors in $\mathbb{R}^n$. Then $E_1,\cdots,E-n$ generate $\mathbb{R}^n$.

Proof. Gievn $X=(x_1,\cdots,x_n)\in\mathbb{R}^n$,
$$X=\sum_{i=1}^nx_iE_i.$$

Proposition. The set of all linear combinations of $v_1,\cdots,v_n$ is a subspace of $V$.

Proof. Straightforward.

Example. Let $v_1$ be a non-zero element of a vector space $V$, and let $w$ be any element of $V$. The set
$$\{w+tv_1: t\in\mathbb{R}\}$$
is the line passing through $w$ in the direction of $v_1$. This line is not a subspace, however if $w=O$, it is a subspace of $V$, generated by a single vector $v_1$.

Example. Let $v_1,v_2$ be two elements of a vector space $V$. The set of all linear combinations of $v_1,v_2$
$$t_1v_1+t_2v_2: t_1,t_2\in\mathbb{R}\}$$
is a plane through the origin and it is a subspace of $V$, generated by $v_1,v_2$. The plane passing through a point $P\in V$, parallel to $v_1,v_2$ is the set
$$\{P+t_1v_1+t_2v_2: t_1,t_2\in\mathbb{R}\}.$$
However, this is not a subspace unless $P=O$.