What do I use to find the image and kernel of a given matrix? I had a couple of questions about a matrix problem. What I'm given is:

Consider a linear transformation $T: \mathbb R^5 \to \mathbb R^4$ defined by $T( \vec{x} )=A\vec{x}$, where
  $$A = \left(\begin{array}{crc}
 1 &  2 &  2 & -5 &  6\\
-1 & -2 & -1 &  1 & -1\\
 4 &  8 &  5 & -8 &  9\\
 3 &  6 &  1 &  5 & -7
\end{array}\right)$$
  
  
*
  
*Find $\mathrm{im}(T)$
  
*Find $\ker(T)$

My questions are:
What do they mean by the transformation?
What do I use to actually find the image and kernel, and how do I do that?

After a long night of studying I finally figured out the answer to these. The previous answers on transformation were all good, but I have the outlined steps on how to find $\mathrm{im}(T)$ and $\ker(T)$.
$$A = \left(\begin{array}{crc}
 1 &  2 &  2 & -5 &  6\\
-1 & -2 & -1 &  1 & -1\\
 4 &  8 &  5 & -8 &  9\\
 3 &  6 &  1 &  5 & -7
\end{array}\right)$$
(1) Find $\mathrm{im}(T)$
$\mathrm{im}(T)$ is the same thing as column space or $C(A)$. The first step to getting that is to take the Transpose of $A$.
$$ 
A^T = \left(\begin{array}{crc}
 1 & -1 &  4 &  3 \\
 2 & -2 &  8 &  6 \\
 2 & -1 &  5 &  1 \\
-5 &  1 & -8 &  5 \\
 6 & -1 &  9 &  -7 
\end{array}\right)$$
once that's done the next step is to reduce $A^T$ to Reduced Row Echelon Form
$$ 
\mathrm{rref}(A^T) = \left(\begin{array}{crc}
 1 & 0 &  1 & -2 \\
 0 & 1 & -3 & -5 \\
 0 & 0 &  0 &  0 \\
 0 & 0 &  0 &  0 \\
 0 & 0 &  0 &  0 
\end{array}\right)$$
now on this step I honestly don't know the reasons behind it, but the thext thing you do is take the rows and that's your answer. so that: 
$$\mathrm{im}(T)\ = \begin{align*}
\operatorname{span}\left\{\left(\begin{array}{crc}
 1 \\
 0 \\
 1 \\
-2 \end{array}\right), \left(\begin{array}{crc}
 0 \\
 1 \\
 -3 \\
-5 \end{array}\right)\right\}
\end{align*}$$
(2) Find $\ker(T)$
$\ker(T)$ ends up being the same as the null space of matrix, and we find it by first taking the Reduced Row Echelon Form of A
$$ 
\mathrm{rref}(A) = \left(\begin{array}{crc}
 1 & 2 &  0 &  3 & -4\\
 0 & 0 &  1 & -4 &  5\\
 0 & 0 &  0 &  0 &  0\\
 0 & 0 &  0 &  0 &  0
\end{array}\right)$$
we then use that to solve for the values of $\mathbb R^5$ so that we get
$$\begin{align*}
\left(\begin{array}{crc}
 x_1 \\
 x_2 \\
 x_3 \\
 x_4 \\
 x_5 \end{array}\right) = r\left(\begin{array}{crc}
 -2 \\
  1 \\
  0 \\
  0 \\
  0 \end{array}\right) + s\left(\begin{array}{crc}
-3 \\
 0 \\
 4 \\
 1 \\
 0 \end{array}\right) + t\left(\begin{array}{crc}
 4 \\
 0 \\
-5 \\
 0 \\
 1 \end{array}\right)
\end{align*}$$
from that we arrange the vectors and get our answer the vectors and that gives us our answer 
$$\begin{align*}
\ker(T) = \operatorname{span}\left\{\left(\begin{array}{crc}
-2 \\
 1 \\
 0 \\
 0 \\
 0 \end{array}\right), \left(\begin{array}{crc}
-3 \\
 0 \\
 4 \\
 1 \\
 0 \end{array}\right), \left(\begin{array}{crc}
 4 \\
 0 \\
-5 \\
 0 \\
 1 \end{array}\right)\right\}
\end{align*}$$
and that's that.
