Basis of Kernel of a matrix

Question

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.

Answer 1

recall rank-nulity theorem namely

$rank(A)+rank(nullspace)=n$

in your case $n=6$ ,and what you need it to find rank of given matrix

for find basis of nullspace,this may help you

Finding the basis of a null space

Answer 2

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: $$\mathrm{span}\{(1,1,1,1,\theta,\theta)\}.$$