Show that the kernel ker(B) is a vector subspace of the domain. I want to know how to show that $\ker(B)$ is a vector subspace of the domain.
 A: Theorem: the kernel of any linear map $f : V \to W$ is a vector subspace of the domain. Proof: we must show that $\ker(f) = \{x \in V : f(x) = 0\}$ is closed under zero, addition, and multiplication. Since $f$ is linear, we have $f(0) = 0$, so $0 \in \ker(f)$. Furthermore, if we have $a, b \in \ker(f)$, then $f(a + b) = f(a) + f(b) = 0 + 0 = 0$; then $a + b \in ker(f)$. Finally, if we have scalar $s$ and $a \in \ker(f)$, then $f(sa) = s f(a) = s 0 = 0$; then $sa \in\ker(f)$. Then $\ker(f)$ is a subspace of $V$.
This shows that the kernel of any linear map (and in particular of $B$) is a sub space of the domain.
To find a basis for the kernel, you should use RREF. This is a well-known method and should be in basically any introductory textbook on linear algebra.
A: Any transformation which can be represented by a matrix is linear.  The kernel of any linear transformation is a vector subspace of the domain.  By linearity the kernel satisfies the subspace criterion: closure under addition and scalar multiplication.
You can use Gaussian elimination and back substitution to get a basis.
So playing off what @Bernard did, we get $\{(1,1,0,0,0,0,0),(-2,0,1,1,0,0,0),(-1,0,-3,0,1,1,0)\}$ by plugging in the standard basis for $\Bbb K^3$.
A: I'll sketch how to obtain  the reduced row echelon form:
\begin{align}
&\phantom{{}\rightsquigarrow{}} \begin{pmatrix}
3 & -3 & 1 & 5 & 1 & 5 & 5\\
1 & -1 & 1 & 1 & 1 & 3 & -1\\
2 & -2 & 1 & 3 & 0 & 5 & 2\\
2 & -2 & 0 & 4 & 0 & 2 & 1
\end{pmatrix}\rightsquigarrow
 \begin{pmatrix}
1 & -1 & 1 & 1 & 1 & 3 & -1\\
3 & -3 & 1 & 5 & 1 & 5 & 5\\
2 & -2 & 1 & 3 & 0 & 5 & 2\\
2 & -2 & 0 & 4 & 0 & 2 & 1
\end{pmatrix} \\
&\rightsquigarrow 
 \begin{pmatrix}
1 & -1 & 1 & 1 & 1 & 3 & -1\\
0 & 1 & -2 & 2 & -2 & -4 & 8\\
0 & 0 &- 1 & 1 & -2 & -1 & 4\\
0 & 0 & -2 & 2 & -2 & -4 & 3
\end{pmatrix} \rightsquigarrow 
 \begin{pmatrix}
1 & -1 & 1 & 1 & 1 & 3 & -1\\
0 & 1 & 0 & 0 & 0 & -2 & 0\\
0 & 0 &- 1 & 1 & -2 & -1 & 4\\
0 & 0 & 0 & 0 & 2 & -2 & -5
\end{pmatrix}
\end{align}
You should ultimately obtain, if I'm not mistaken:
$$ \begin{pmatrix}
1 & -1 & 0 & 2 & 0 & 1 & 0\\
0 & 0 & 1 & -1 & 0 & 3 & 0\\
0 & 0 & 0 & 0 & 1 & -1 & 0\\
 0 & 0 & 0 & 0 & 0 & 0 & 1
\end{pmatrix}$$
which shows that
\begin{cases}
x_1=x_2-2x_4-x_6, \\ x_3=x_4-3x_6,\\x_5=x_6,&x_7=0.
\end{cases}
This defines an isomorphism from $K^3$ ($K$ denotes the base field) onto $\ker B$:
\begin{align}
K^3&\xrightarrow{\quad f\quad} \ker B \\
(x,y,z)\:&|\mkern-7mu\xrightarrow{\quad \enspace\quad}(x-2y-z, x, y-3z, y, z,z,0)
\end{align}
and this isomorphism maps any basis of $K^3$, for instance the canonical basis, onto a basis of $\ker B$.
