4.1 Vector Space and Subspace
Vector Space is a set of vectors whose linear combinations belong to the same set. This means that whenever we pick $2$ vectors from a vector space and find thier linear combination, the resultant vector will be in the vector space. A vector space in two and three dimensions is $R^2$ and $R^3$.
We can make some common observations about vector spaces. Let $u,v$ be two vectors, then their linear combination can be described as $w=au+bv$ where $a,b$ are scalars. If $a=b=0$, $w=au+bv=0$. This means that for any vector space (which contains vectors $u,v$), the zero vector will always be in it. This observation gives us a powerful tool to deduce whether a set is a vector space or not. If zero vector is not in the set, it’s not a vector space.
A vector space that is contained within another vector space is called a subspace of that space. A space defined in $R^2$ contianing just zero vector as $S=\bigg\{\begin{bmatrix} 0 \\ 0 \end{bmatrix}\bigg\}$ is a subspace as any linear combination of zero vector is still a zero vector. Another example of subspace in $R^2$ is a line passing through origin as any multiple of the vector will also be on the line and hence fulfilling the criteria of a vector space. Any plane passing through origin is a subspace in $R^3$ as linear combination of the vectors in the plane will still reside in the plane.
4.2 Union and Intersection of Vector Space
Let $S_1,S_2$ be two vector spaces, their union $S_1 \cup S_2$ is not a subsapce. For example, let $S_1$ and $S_2$ be two lines, the sum of two non-zero vectors in $S_1$ and $S_2$ respectively will not be on any of the lines and hence the union is not a subspace.
The intersection of the subspaces denoted as $S_1 \cap S_2$ is a subspace. Let there are two vectors $u,v$ such that both $u,v \in S_1 \cap S_2$. This means that $u,v \in S_1$ and $u,v \in S_2$. For $S_1 \cap S_2$ to be a subsapce, $au + bv \in S_1 \cap S_2$ where $a$ and $b$ are scalars. As $u,v \in S_1$ and $S_1$ is a subspace, it’s linear combination $au + bv \in S_1$. Similarly, $au + bv \in S_2$ and hence $au + bv \in S_1 \cap S_2$.
4.3 Column Space
Given a matrix $A$, it’s columns and all their linear combinations form a vector space called as the column space $C(A)$ of $A$. For example, for the matrix $A = \begin{bmatrix} 1 & 3 \\ 2 & 3 \\ 4 & 1 \end{bmatrix}$ it’s column space $C(A)$ is a plane passing through $\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$ (origin), $\begin{bmatrix} 1 \\ 2 \\ 4 \end{bmatrix}$, $\begin{bmatrix} 3 \\ 3 \\ 1 \end{bmatrix}$ in $R^3$.
Let us look at another example. Following matrix $A$ is in $R^4$. The column space $C(A)$ of $A$ is formed by all linear combinations of columns of $A$. Let there be a system of linear equations $Ax = b$ where $x,b$ are column vectors. Does this system of linear equation will have a solution $x$ for all values of $b$?
$$\begin{align} A = \begin{bmatrix} 1 & 1 & 2 \\ 2 & 1 & 3 \\ 3 & 1 & 4 \\ 4 & 1 & 5 \end{bmatrix}, x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}, b = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \\ b_4 \end{bmatrix} \end{align}$$
The system of linear equation $Ax = b$ can be further reduced to:
$$\begin{align} Ax =b \implies\begin{bmatrix} 1 & 1 & 2 \\ 2 & 1 & 3 \\ 3 & 1 & 4 \\ 4 & 1 & 5 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = b \implies x_1\begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix} + x_2\begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix} + x_3\begin{bmatrix} 2 \\ 3 \\ 4 \\ 5 \end{bmatrix} = b \end{align}$$
This means that for $Ax=b$ to have a solution, $b$ should be the linear combination of the columns of $A$. Hence, the system of linear equation $Ax=b$ will have a solution only if $b \in C(A)$, where $C(A)$ is the column space of $A$.
If we further look at the columns of $A$, $col_3$ is a linear combination of $col_1$ and $col_2$ as $col_3 = col_1 + col_2$. This means that $col_3$ does not contribnute or add any dimension while forming the column space of $A$. $C(A)$ can be defined just using $col_1$ and $col_2$. Hence column space $C(A)$ of $A$ is a two dimensional plane in $R^4$.
4.4 Null Space
Null space of marix $A$ contains all solutions of $Ax=0$, i.e null space is the solution of following equation for above $A$. Here the null space will be in $R^3$ as $x$ is a three dimensional vector.
$$\begin{align} Ax=0 \implies \begin{bmatrix} 1 & 1 & 2 \\ 2 & 1 & 3 \\ 3 & 1 & 4 \\ 4 & 1 & 5 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} \end{align}$$
Some the vectors which will be in the null space of $A$ are: $\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix}, \begin{bmatrix} 2 \\ 2 \\ -2 \end{bmatrix},…$ In general, all vectors of the form $\begin{bmatrix} c \\ c \\ -c \end{bmatrix}$ forms the null space of $A$.
For these set of vectors to be a subspace, for two solutions $u,v$ of $Ax=0$, their linear combination $au+bv$ should also be a solution of $Ax=0$. As $u, v$ are solutions of $Ax=0$, $Au=0;Av=0$. This means:
$$\begin{align} aAu=0;bAv=0 \implies Aau=0; Abv=0 \implies Aau+Abv=0 \implies A(au+bv)=0 \end{align}$$
Hence, $au+bv$ is also a solution of $Ax=0$ which makes nullspace a vector space. One of the major intution about null space is the fact that for any matrix $A$, null space will contain non-zero vectors only if all of it’s columns are not lineraly independent, i.e. at least one of the columns of $A$ should be the linear combination of all the other columns.
Another important question which can be asked that whether for any non-zero $b$, whether the solutions of the equation $Ax=b$ forms a subspace. No, they don’t. As for non-zero $b$, the zero vector is not a solution and hence the set of solutions don’t form a subspace as every subspace should contain the zero vector.