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?
 A: What they mean by the transformation $T$ is the transformation which is induced by multiplication by $A$. You can verify that matrix multiplication is in fact a linear mapping, and in our particular case we have the linear mapping $T:\ \mathbf{x}\mapsto A\mathbf{x}$.
The image is then defined as the set of all outputs of the linear mapping. That is
$$\operatorname{Im}(T) = \left\{\mathbf{y}\in \mathbb{R}^4\ \big|\ \mathbf{y} = A\mathbf{x}\ \text{such that}\ \mathbf{x}\in\mathbb{R}^5 \right\}$$
If you play around with the mapping a little bit then you should find that the image is in fact a very familiar subspace associated with the matrix $A$ (take a look at how the mapping $T$ acts on the standard basis).
The kernel is correspondingly defined as the set of all inputs which are taken to zero.
$$\ker(T) = \left\{\mathbf{x}\in \mathbb{R}^5\ \big|\ A\mathbf{x} = \mathbf{0} \right\}$$
Again, there is a familiar subspace of the matrix $A$ associated with the kernel, look carefully at the definition and you should be able to figure out what it is.
A: 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.
A: By a linear transformation, they mean a function between vector spaces which satisfies $T(cx + y) = cT(x) + T(y)$. In our case, this transformation is multiplication by the matrix $A$. 
The image is the set of all points in $\mathbb{R}^4$ that you get by multiplying this matrix to points in $\mathbb{R}^5$, you can find these by checking the matrix on the standard basis.
The kernel is the set of all points in $\mathbb{R}^5$ such that, multiplying this matrix with them gives the zero vector. Again you can find this in a similar way.
A: I could give an explanation for the most appreciated answer why image is calculated in this way. Image of a matrix is basically all the vectors you can obtain after this linear transformation. Let's say $A$ is a $2 \times 2$ matrix
$$A=\pmatrix {a_1 & b_1\\ a_2 & b_2}$$
. If we apply A as a linear transformation to the standard base, aka the identity matrix, we get A itself. However, we could consider this transformation as it transforms the basis vectors to all the columns A has. (1, 0) to (a1, a2), (0, 1) to (b1, b2). Therefore, the image of A is just the span of the basis vectors after this linear transformation; in this case, span ((a1, a2), (b1, b2)). This is the reason why we need to get rref of the transpose of A. We are simplely getting linearly independent basis vectors after this linear transformation. If there's anything unclear, I really recommand you to watch this video made by 3Blue1Brown, it shows this in a visual way. Here's the link: https://www.youtube.com/watch?v=uQhTuRlWMxw
