Given a matrix of basis transformation what is the algorithm to find $ker(T)$ and $im(T)$? I'm given the following transformation matrix of the linear map $T:\mathbb R^4\to\mathbb R^3$:

Find  $\mathrm{ker}(T)$ and $\mathrm{im}(T)$
So I should probably get this matrix to $\mathrm{rref}$:

But then what ? Also, what does the $\mathrm{rref}$ mean for this matrix ?
 A: Here's one way to think about it: when we row reduce a matrix, what we're doing is computing another matrix by matrix multiplication on the left by some invertible matrix.  That is starting with are matrix $A$, we have
$$
E\cdot A = E_n\cdots E_1 \cdot A = \pmatrix{1 & 0 & 6 & 6\\ 0 & 1 & -3/2 & -1\\ 0&0&0&0}
$$
Where each of $E_1,\dots,E_n$ is an elementary matrix.  Because $E$ is invertible, this product has the following essential property:

For a vector $x$, $EAx = 0$ if and only if $Ax = 0$

We may deduce two things from the above: first, the kernel of $EA$ (the reduced matrix) is the same as the kernel as $A$ (it is easier, however, to directly find the kernel of $EA$, which is in reduced row echelon form).  Second, the image of $A$ is $E^{-1}$ applied to the vectors in the image of $EA$.  What this means is that the image of $A$ will be the span of the columns in $A$ corresponding to the position of the pivots in rref matrix $EA$.  So, in this example, the image of $A$ is the span of the vectors
$$
\pmatrix{1\\1\\2},\pmatrix{2\\4\\2}
$$
Since $EA$ has a pivot ("leading $1$") in its first column and in its second. 
Let me know if there's anything here you'd like me to clarify.  Hope that helps.

Elaboration regarding the image of $A$:
The definition of the image of $A$ is the set of all vectors $v$ for which $v = Ax$ for some vector $x$ in the domain of $A$.  The image of any matrix $A$ will necessarily be the span of all of its column vectors.  So, in this case, the Image of $A$ will be the span of the vectors
$$
\pmatrix{1\\1\\2},\pmatrix{2\\4\\2}, \pmatrix{3\\0\\9}, \pmatrix{4\\2\\10}
$$
The reason that this is the case is that we can, using matrix multiplication, write
$$
A \cdot \pmatrix{a\\b\\c\\d} = 
a\pmatrix{1\\1\\2} + b\pmatrix{2\\4\\2} + c\pmatrix{3\\0\\9} + d\pmatrix{4\\2\\10}
$$
While we can describe any image in these terms, it is not generally ideal to do so since the column vectors of a matrix are not generally linearly independent (as is the case for our $A$), and thus they will not form a basis of the image.  In order to find a basis, we can take a smallest linearly independent subset of these vectors.
Now, by our previous statements, the image of $EA$ is the span of the vectors
$$
\pmatrix{1\\0\\0},\pmatrix{0\\1\\0},\pmatrix{6\\-3/2\\0}, \pmatrix{6\\-1\\0}
$$
However, in this case, it's clear that we can find a basis of the image of $EA$ by simply taking the vectors corresponding to the pivots.  That is, the image of $EA$ is the span of the  vectors
$$
\pmatrix{1\\0\\0},\pmatrix{0\\1\\0}
$$
Which are, in fact, linearly independent.  What this means then is that for any vector $x$, we have
$$
EAx = a\pmatrix{1\\0\\0} + b\pmatrix{0\\1\\0} = \pmatrix{a\\b\\0}
$$
For some choice of $a,b \in \mathbb{R}$. Now, note that
$$
\pmatrix{1\\0\\0} 
= EA\cdot \pmatrix{1\\0\\0\\0} 
= E\left(A\cdot \pmatrix{1\\0\\0\\0}\right)
= E \cdot \pmatrix{1\\1\\2}\\
\pmatrix{0\\1\\0} 
= EA\cdot \pmatrix{0\\1\\0\\0} 
= E\left(A\cdot \pmatrix{0\\1\\0\\0}\right)
= E \cdot \pmatrix{2\\4\\2}
$$
It follows that
$$
Ax = E^{-1}(EAx) = E^{-1}\cdot \left(a\pmatrix{1\\0\\0} + b\pmatrix{0\\1\\0}\right) 
= a\pmatrix{1\\1\\2} + b\pmatrix{2\\4\\2}
$$
So that these two vectors form a basis of the image, as desired.
A: Think of what you did as the coefficients' matrix of a homogeneous system with four variable $\;x_i\;,\;i=1,2,3,4\;$ in order to find the kernel, so what you're said there is:
$$\text{Second row:}\;\;x_2-\frac32x_3-x_4=0\iff 2x_2=3x_3+2x_4\\
\text{First row:}\;\;x_1+6x_3+6x_4=0\implies x_1=-6x_3-6x_4$$
and you have two free variables $\;x_3,x_4\;$ (and thus its dimension's two), so a basis for the kernel is
$$\left\{\;x_3=1\,,\,x_4=0\implies\begin{pmatrix}\!\!-6\\3\\1\\0\end{pmatrix}\;\;,\;\;x_3=0\,,\,x_4=1\implies\begin{pmatrix}\!\!-6\\2\\0\\1\end{pmatrix}\;\right\}$$
Do now something similar with the original matrix's columns, and take into account that in this case $\;\dim\text{Im}\,T=2\;$ (why?)
