Find the kernel and the image of the following vector? Find the Kernel and the image of the following matrix:
$\begin{bmatrix}1 & 1 & 1\\1 & -1 &1\end{bmatrix}$
In order to find the kernel I must find the basis for the augmented matrix with a column of zeroes, correct? So I row reduced and then got $x_1+x_2+x_3=0$ and $2x_2=0$ which means$ x_1=-x_3, x_2=0, x_3=-x_1$ then I wrote my answer as $x_1$$\begin{bmatrix}0\\0\\-1\end{bmatrix}$+$x_3$$\begin{bmatrix}-1\\0\\0\end{bmatrix}$
 Is this correct for the kernel? and how do I find the image? 
 A: The augmented matrix will be \begin{bmatrix}1 & 1 & 1 & 0 \\ 1 & -1 & 1 & 0\end{bmatrix} and reduced to \begin{bmatrix}1 & 0 & 1 & 0\\0 & 1 & 0 & 0\end{bmatrix}
Let $x_1=t$, then
$$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}=t\begin{bmatrix}1 \\ 0 \\ -1\end{bmatrix}$$
Thus the kernel of $A$ is $\operatorname{span}([1,0,-1]^{T})$.
To find the image of $A$, set the augmented matrix\begin{bmatrix}1 & 1 & 1 & y_1\\1 & -1 & 1 & y_2\end{bmatrix}
then it is reduced to\begin{bmatrix}1 & 0 & 1 & \frac{y_1+y_2}{2}\\ 0 & 1 & 0 & \frac{y_1-y_2}{2}\end{bmatrix}
and
$$
\begin{bmatrix}\frac{y_1+y_2}{2} \\ \frac{y_1-y_2}{2}\end{bmatrix}=\frac{y_1}{2}\begin{bmatrix}1\\1\end{bmatrix}+\frac{y_2}{2}\begin{bmatrix}1 \\ -1\end{bmatrix}
$$
Therefore, the image of $A$ is $\operatorname{span}([1,1]^{T},[1,-1]^{T})$.
A: You find that $x_3$ is a free variable.  Letting $x_3$ be parametrized as $x_3=t$, we have the system $$\begin{cases} x_1=-t\\x_2=0\\x_3=t\end{cases}$$
In other words, $\begin{bmatrix} x_1\\x_2\\x_3\end{bmatrix} = \begin{bmatrix}-t\\0\\t\end{bmatrix} = t\cdot \begin{bmatrix}-1\\0\\1\end{bmatrix}$
Your proposed kernel does not make use of the fact that $x_1$ is dependent on the value of $x_3$ and vice versa, choosing to instead write them as separate pieces implying that either could be any value without regards to the value of the other.  Note that setting $x_1=0$ and $x_3=-1$ would have implied that the vector $[1~~0~~0]^T$ would be in the kernel, but applying the multiplication it is clearly not.
The image can be found a number of ways, but since you have already found the reduced row echelon form of the matrix, you can note that the locations of the columns with pivots in the rref of the matrix will correspond to locations of columns in the original matrix which form a maximal independent set which form a basis for the image (columnspace).
In your case, you found a pivot in the first and second column, implying that $\begin{bmatrix}1\\1\end{bmatrix}$ and $\begin{bmatrix}1\\-1\end{bmatrix}$ form a basis for your image.  If you so choose, you can simplify how this appears applying a process such as gram-schmidt orthogonalization (which is usually taught after one is more comfortable with finding kernels and images as well as the foundations of inner-product spaces are conveyed).  Or, you may choose to note that as it is a subspace of dimension two of $\Bbb R^2$, it must be all of $\Bbb R^2$.
Either way, one can see that the image is in fact $\Bbb R^2$, i.e. $span\left(\begin{bmatrix}1\\0\end{bmatrix},\begin{bmatrix}0\\1\end{bmatrix}\right)$
