Kernel Explanation

Question

sorry for asking so many questions lately but our lecturer is doing a terrible job explaining things.

Calculate $ker(A)$ given that:

$f:\{\mathbb{R}^3→\mathbb{R}^3; r→ A\vec{r}\}$

$A= \bigl(\begin{smallmatrix} 1&2 &4 \\ 0&1 &2 \\ 3&1 &2 \end{smallmatrix}\bigr)$

I have been browsing the web for some answers but I don't really get it. Maybe someone can explain what the kernel is and how I calculate it. I hope my question makes sense. I had to translate it since I am studying at a german university.

Thanks in advance

Answer 1

Solve the homogeneous system, which means reducing the matrix by rows, say:

$$\begin{pmatrix}1&2&4\\ 0&1&2\\3&1&2\end{pmatrix}\stackrel{R_3-3R_1}\longrightarrow\begin{pmatrix}1&2&4\\ 0&1&2\\0&\!\!-5&\!\!-10\end{pmatrix}\stackrel{R_3+5R_2}\longrightarrow\begin{pmatrix}1&2&4\\ 0&1&2\\0&0&0\end{pmatrix}$$

The above means

$$\begin{align}&x_2+2x_3=0\implies x_2=-2x_3\\ &x_1+2x_2+4x_3=0\implies x_1=-2(-2x_3)-4x_3=0\end{align}$$

and thus the kernel is

$$\left\{\;\begin{pmatrix}0\\-2t\\t\end{pmatrix}\;\;:\;\;\;t\in\Bbb R\;\right\}$$

Answer 2

The kernel is the set $\{\vec r : A \vec r = \vec 0\}$. You can determine all solutions using e.g. row reduction.

Answer 3

The kernel of a linear transformation is the set of vectors that it sends to 0. Thus the goal is to find all solutions $v$ of $$Av=0$$ This can be done with Gaussian elimination.