Determining the null space of the matrix 
Determine the null space of the matrix:$$\begin{bmatrix} 1 & -1 \\ 2 & 3 \\ 1 & 1 \end{bmatrix}$$

My try:
$$\begin{bmatrix} 1 & -1 \\ 2 & 3 \\ 1 & 1 \end{bmatrix}_{R_2\rightarrow R_2-2R_1\\R_3\rightarrow R_3-R_1}$$
$$\begin{bmatrix} 1 & -1 \\ 0 & 5 \\ 0 & 2 \end{bmatrix}_{R_3\rightarrow 5R_3-2R_2}$$
$$\begin{bmatrix} 1 & -1 \\ 0 & 5 \\ 0 & 0 \end{bmatrix}_{R_2\rightarrow \frac{R_2}{5}}$$
$$\begin{bmatrix} 1 & -1 \\ 0 & 1 \\ 0 & 0 \end{bmatrix}$$
From this I got $$x-y=0\implies x=y\\y=0$$
$$(x,y,z)^T=(y,0,z)^T=y(1,0,0)^T+z(0,0,1)^T$$
So, $(1,0,0)^T$ and $(0,0,1)^T$ is the null space. Is this correct?
 A: Recall that by definition the nullspace is the subspace of all vectors $\vec x$ such that $A\vec x=\vec 0$ and in that case we have ony the trivial solution $(x_1,x_2)=(0,0)$ then $Null(A)=\{\vec 0\}$.
Notably to solve $Ax=0$ we can proceed by RREF to obtain
$$\begin{bmatrix} 1 & -1 \\ 2 & 3 \\ 1 & 1 \end{bmatrix}\to 
\begin{bmatrix} 1 & -1 \\ 0 & 5 \\ 0 & 2 \end{bmatrix}\to 
\begin{bmatrix} 1 & -1 \\ 0 & 1 \\ 0 & 0 \end{bmatrix}$$
that is
$$\begin{bmatrix} 1 & -1 \\ 0 & 1 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \end{bmatrix}$$
that is $x_1=x_2=0$.
A: Right Nullspace
If vector $\alpha = [x,y]$ is in the right nullspace of A then 
$$\begin{bmatrix}
1 & -1 \\
2 & 3 \\
1 & 1 \\
\end{bmatrix}[x,y]^T = \begin{bmatrix} 0 \\ 0 \\ 0
\end{bmatrix}$$
This gives
$$x - y = 0$$
$$2x + 3y = 0$$
$$x + y = 0$$
First and third equation tells us that $x = y = -x$, i.e. the only solution is $[0,0]$. So, the right nullspace has dimension zero 
Left Nullspace
If vector $\alpha = [x,y,z]$ is in the left nullspace of A then 
$$\begin{bmatrix}
x & y & z 
\end{bmatrix}
\begin{bmatrix}
1 & -1 \\
2 & 3 \\
1 & 1 \\
\end{bmatrix} = \begin{bmatrix}
0 & 0
\end{bmatrix}$$
So we get 
$$x + 2y + z = 0$$
and 
$$-x + 3y + z = 0$$
First equation tells us $$x = -(2y+z)$$
and second one says
$$x = 3y+z$$
So by equating $x=x$, we get
$$-2y-z = 3y+z$$
that is 
$$5y = -2z$$
or $$y = - \frac{2}{5}z$$. Plugging this back into one of the two x equations, we get
$$x= 3y + z = 3(- \frac{2}{5}z) + z = \frac{-6+5}{5}z = -\frac{1}{5}z$$
So, the Null space takes the form
$$\begin{bmatrix}
-\frac{1}{5} \\
- \frac{2}{5} \\
1
\end{bmatrix}
z$$
A: Your method is correct to find that $x=y$ and $y=0$. But you have made an incorrect conclusion after that stage and you have made a mistake in the dimension of your vectors.
Firstly, see that your matrix acts on vectors in $\mathbb{R}^2$ to form vectors in $\mathbb{R}^3$ like so:
$$\left( \begin{matrix} 1 & -1 \\ 2 & 3 \\ 1 & 1 \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right)=\left( \begin{matrix} x-y\\ 2x+3y\\ x+y \end{matrix} \right)
$$
So you're searching for vectors $(x,y)^T$ in your null space.
If $x=y$ and $y=0$ then $x=y=0$, showing that your null space is just $\{(0,0)^T\}$
