Basis of Kernel of a matrix Given $\theta>0$. Let $H$ be $5 \times 6$ matrix
$$\left[\begin{matrix}
  1 & -1 & 0 & 0 & 0 & 0 \\
  1 & 0 & -1 & 0 & 0 & 0 \\
 1 & 0 & 0 & -1 & 0 & 0 \\
 0 & 0 & 0 & 0 & 1 & -1 \\
 \theta & 0 & 0 & 0 & 0 & -1
 \end{matrix}\right]$$
Consider the subspace $S=\{x\in\mathbb{R}^6:Hx=0$}. I know the subspace $S$ has dimension $1$. However, I couldn't find the basis of $S$. Could anybody help me please. Thanks in advance.

We perform the row operations:
$$\begin{bmatrix}
1 & -1 & 0 & 0 & 0 & 0 \\
1 & 0 & -1 & 0 & 0 & 0 \\
1 & 0 & 0 & -1 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & -1 \\
\theta & 0 & 0 & 0 & 0 & -1 \\
\end{bmatrix}
\xrightarrow{R_5 \gets \tfrac{1}{\theta} R_5}
\begin{bmatrix}
1 & -1 & 0 & 0 & 0 & 0 \\
1 & 0 & -1 & 0 & 0 & 0 \\
1 & 0 & 0 & -1 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & -1 \\
1 & 0 & 0 & 0 & 0 & -\tfrac{1}{\theta} \\
\end{bmatrix}
\xrightarrow{R_i \gets R_{\alpha(i)} \text{ where } \alpha=(15432)}
\begin{bmatrix}
1 & 0 & 0 & 0 & 0 & -\tfrac{1}{\theta} \\
1 & -1 & 0 & 0 & 0 & 0 \\
1 & 0 & -1 & 0 & 0 & 0 \\
1 & 0 & 0 & -1 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & -1 \\
\end{bmatrix}
\xrightarrow{R_i \gets R_i-R_1 \text{ for } i \in \{2,3,4\}}
\begin{bmatrix}
1 & 0 & 0 & 0 & 0 & -\tfrac{1}{\theta} \\
0 & -1 & 0 & 0 & 0 & \tfrac{1}{\theta} \\
0 & 0 & -1 & 0 & 0 & \tfrac{1}{\theta} \\
0 & 0 & 0 & -1 & 0 & \tfrac{1}{\theta} \\
0 & 0 & 0 & 0 & 1 & -1 \\
\end{bmatrix}
\xrightarrow{R_i \gets -R_i \text{ for } i \in \{2,3,4\}}
\begin{bmatrix}
1 & 0 & 0 & 0 & 0 & -\tfrac{1}{\theta} \\
0 & 1 & 0 & 0 & 0 & -\tfrac{1}{\theta} \\
0 & 0 & 1 & 0 & 0 & -\tfrac{1}{\theta} \\
0 & 0 & 0 & 1 & 0 & -\tfrac{1}{\theta} \\
0 & 0 & 0 & 0 & 1 & -1 \\
\end{bmatrix}
$$
which is in reduced row echelon form.  So the matrix has rank $5$ and hence indeed the nullity is $6-5=1$, by the Rank-Nullity Theorem.
By inspection the null space is: $$\mathrm{span}\{(1,1,1,1,\theta,\theta)\}.$$
