Finding the basis of the kernel of a specific linear mapping (check) I am asked to find the basis of the kernel of $\beta$, where $\beta$: $\mathbb{R^3}$ $\rightarrow$ $\mathbb{R^3}$,
$$ \beta (x,y,z) =
        \begin{pmatrix}
        x+y \\
        2x+y-z \\
        x-z \\
        \end{pmatrix}
$$
I've put the matrix 
$$ \beta (x,y,z) =
        \begin{pmatrix}
        1&1&0 \\
        2&1&-1\\
        1&0&-1 \\
        \end{pmatrix}
$$
into row reduced echelon form and ended up with the standard 3x3 identity matrix. Is the basis therefore 
$$ \beta (x,y,z) =
        \begin{pmatrix}
        1 \\
        1\\
        1 \\
        \end{pmatrix}
$$
or have I done this incorrectly? I would say this is a simple example but the answer I have is different to the one given which is $(1,-1,1)$. Thanks in advance.
 A: First of all, if you indeed end up with identity matrix, it means that $\beta$ is of full rank, i.e. isomorphism, which would mean that the kernel is trivial. This makes your conclusion invalid.
What you have to do is solve linear equation $\beta x = 0$. When you reduce the matrix you finally get 
\begin{pmatrix}
1 & 0 & -1\\
0& 1 & 1\\
0 & 0 & 0
\end{pmatrix}
So, the matrix is actually of rank $2$, which means that kernel is one-dimensional. It is spanned by $(1,-1,1)$ because, as you can check, it indeed solves the system with above matrix. What we did is look at last column, multiplied it by $-1$ and replaced $0$ with $1$ (standard procedure when solving linear systems).
A: Row reducing the matrix should've given you the matrix
$$         \begin{pmatrix}
        1&1&0 \\
        0&1&1 \\
        0&0&0\\
        \end{pmatrix}
$$
rather than the identity matrix, since the determinant of the matrix is zero: the second row is the sum of the first and third rows!
It is then easy to see that a vector $(x,y,z)$ is an element of the kernel if and only if  $x+y = 0$ and $y+z = 0$.
From these equations, you can derive that $$\operatorname{ker}(\beta) = \{(x,-x,x) \mid x \in \mathbb{R}\},$$
which is generated by the vector $(1,-1,1)$ as per your solutions.
