I'm trying to find the $$
ker\begin{bmatrix}
0 & 0 & 0 \\
0 & 1 & 2 \\
0 & 2 & 4 \\
\end{bmatrix}
$$
I would usually write the linearly dependent columns in terms of the others. For example: $$ v_1+2v_2-v_3=0 $$
In this case I'm not sure what to do with column $ v_1 $ which consists entirely of $0$s. It would give me a basis that looks like
$$ \begin{bmatrix}
1 \\
2 \\
-1 \\
\end{bmatrix} ,
\begin{bmatrix}
2 \\
2 \\
-1 \\
\end{bmatrix},…,
\begin{bmatrix}
a \\
2 \\
-1 \\
\end{bmatrix}
$$
for all $a$.
I know I should have a 2-dimensional basis, though. I suspect I'm wrong in my interpretation of linear (in)dependence? Or am I missing some obvious way to combine these specific columns?
Best Answer
Call the matrix $A$. Then to find $\ker A$, we row-reduce the augmented matrix $[A \mid \vec 0]$ to obtain: $$ \left[ \begin{array}{ccc|c} 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right] $$ Hence, $v_2$ is the basic variable that is to be expressed in terms of the free variables $v_1$ and $v_3$. Since: $$ v_2 + 2v_3 = 0 \iff v_2 = -2v_3 $$ it follows that the desired kernel is the set of all vectors of the form: $$ \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = \begin{bmatrix} v_1 \\ -2v_3 \\ v_3 \end{bmatrix} = v_1\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} + v_3 \begin{bmatrix} 0 \\ -2 \\ 1 \end{bmatrix} $$ where $v_1, v_3$ are any real numbers.