Finding basis for image and kernel of A = [x+y,x-y,x+z,x-z,x+y+z] Here is a problem I'm trying to solve:
Knowing that vectors $\mathbf{x}, \mathbf{y}, \mathbf{z} \in \Bbb{R}^{2013}$ are linearly independent, find basis for image and kernel of
$$A=[\mathbf{x+y},\mathbf{x-y},\mathbf{x+z},\mathbf{x-z},\mathbf{x+y+z}]$$
I've been able to find the basis for image of $A$: The columns in $A$ are linear combinations of $\mathbf{x,y,z}$, ($\mathbf{x}=\frac{1}{2}((\mathbf{x+y})+(\mathbf{x-y}))$, and then $\mathbf{y} = \mathbf{x+y-x}, \mathbf{z}=\mathbf{x+z-x}$, so $im (A) = span (\mathbf{x,y,z})$.
But when it goes to the kernel I have no idea what to do - vectors are abstract so the algorthm doesn't work. All I've came up with is that it's dimension must be $5-rank A=5-3=2$, but what to do next?
 A: $u = [u_1,u_2,u_3,u_4,u_5]^T \in ker(A)$ if and only if
$$ u_1 (\mathbf{x+y}) + u_2 (\mathbf{x-y}) + u_3 (\mathbf{x+z}) + u_4 (\mathbf{x-z})
   +u_5 (\mathbf{x+y+z}) = 0. $$
Now you can rearrange the left hand side by $\mathbf{x}, \mathbf{y}, \mathbf{z}$, i.e.
$$ (u_1+u_2+\dots+u_5)\mathbf{x} + (u_1-u_2+u_5)\mathbf{y} + (u_3-u_4+u_5)\mathbf{z} = 0. $$
As $\mathbf{x}, \mathbf{y}, \mathbf{z}$ are linearly independent, you get
$$ u_1 + \dots + u_5 = 0, \quad u_1-u_2+u_5=0, \quad u_3-u_4+u_5 = 0. $$
Now you need to find a basis for the subspace of $R^5$ defined by this
system of equations.
A: Consider the linear map $f\colon\mathbb{R}^3\to\mathbb{R}^{2013}$ defined by $f(e'_1)=x$, $f(e'_2)=y$ and $f(e'_3)=z$, where $\{e'_1,e'_2,e'_3\}$ is the standard basis of $\mathbb{R}^3$. Then $f$ is injective, because its image has dimension $3$.
Consider also $g\colon\mathbb{R}^5\to\mathbb{R}^3$ defined by
\begin{align}
g(e_1)&=e'_1+e'_2\\
g(e_2)&=e'_1-e'_2\\
g(e_3)&=e'_1+e'_3\\
g(e_4)&=e'_1-e'_3\\
g(e_5)&=e'_1+e'_2+e'_3
\end{align}
where $\{e_1,e_2,e_3,e_4,e_5\}$ is the standard basis of $\mathbb{R}^5$.
Then the matrix of $f\circ g$ with respect to the standard bases on $\mathbb{R}^3$ and $\mathbb{R}^{2013}$ is $A$. Since $f$ is injective, the kernel of $f\circ g$ is the same as the kernel of $g$ and the matrix of $g$ (with respect to the standard bases) is
$$
B=\begin{bmatrix}
1 & 1 & 1 & 1 & 1\\
1 & -1 & 0 & 0 & 1\\
0 & 0 & 1 & -1 & 1
\end{bmatrix}
$$
An easy elimination gives the RREF
$$
U=\begin{bmatrix}
1 & 0 & 0 & 1 & 1/2\\
0 & 1 & 0 & 1 & -1/2\\
0 & 0 & 1 & -1 & 1
\end{bmatrix}
$$
which corresponds to the equations
$$
\begin{cases}
x_1=-x_4-\frac{1}{2}x_5\\
x_2=-x_4+\frac{1}{2}x_5\\
x_3=x_4-x_5
\end{cases}
$$
so a basis for the kernel is
$$
\left\{
\begin{bmatrix}
-1\\
-1\\
1\\
1\\
0
\end{bmatrix},
\begin{bmatrix}
-1/2\\
1/2\\
-1\\
0\\
1
\end{bmatrix}
\right\}
$$
