How to find a basis for the kernel and image of a linear transformation matrix

I´ll give a partial answer for the kernel. The kernel is the set of all $x \in \mathbb{R}^{4}$ such that $T(x) = 0$, that is $Ax = 0$. This is an homogeneus system in four unknowns and three equations. The matrix has rank 1, because there´s only one linearly independent column, so you have only one linearly independent equation. For example: $$ -x_{3}+3x_{4} = 0. $$ Therefore $$ \ker(T) = \{(x_{1}, x_{2}, x_{3}, x_{4}) \in \mathbb{R}^{4}: -x_{3}+3x_{4} = 0\} $$ One linearly independent equation means that the kernel has dimension $4-1 = 3$, yo you need three linearly independent vectors that verify that equation to get a basis of the kernel. For example: $$ \{(1, 0, 0, 0),\ (0, 1, 0, 0),\ (0, 0, 3, 1)\} $$

By elimination, you will end up with $$ \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 3 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix} $$ which gives you the dimension of the image of $T$ and the kernel of $T$.

Now, you can read from the matrix a basis for the image of $T$. On the other hand $$ -x_3+3x_4=0 $$ tells you how to find a basis for the kernel of $T$.

Image is spanned on linearly independent columns of matrix $A$, which follows from definition of matrix of linear transformation. To find it do the row reduction of $A^T$ and its nonzero columns span the image, or equivalently, if you row reduce $A$ then columns with pivots will correspond to columns of $A$ that form image of the transformation. Kernel is set of vectors $x$ such that $Ax=0$. To find it's base you need to solve system of equations given by $Ax=0$.

The kernel basis is:

$$ \left[ \begin{array}{cc} 0&0&1&1/3\\ \end{array} \right]^T, \left[ \begin{array}{cc} 1&0&0&0\\ \end{array} \right]^T, \left[ \begin{array}{cc} 0&1&0&0\\ \end{array} \right]^T$$

The image basis is:

$$\left[ \begin{array}{cc} 6&-1&-2\\ \end{array} \right]^T$$