I've never really had much of an interest in linear algebra because I was never really good at it but recently, as part of my degree I've gotten a better handle on it than before. Currently we're covering linear transformations including the idea of kernels and images. However I'm really confused as to what Kernel actually means and additionally I have the following question where I have to find the Kernel of $V$.
If we have that $V=\left\{{\ \begin {pmatrix} y_1\\ y_2 \\ y_3 \end{pmatrix} | \ V:{R^3} \to {R^3}} \ \right\}$ and goes through a transformation and ends up with $\ \begin {pmatrix} y_2-y_3\\ 0\\ 0\end{pmatrix} $, I need to find $Kern(V)$. To do that I followed the following formula:
$$Kern(V):=\left\{{{\vec v \in {R^3}} \ | L(\vec v)=\vec 0} \ \right\}$$
$$\ \begin {pmatrix} y_2-y_3\\ 0\\ 0\end{pmatrix} =\ \begin {pmatrix} 0\\ 0\\ 0\end{pmatrix}$$ $$y_2=y_3≠0$$
But I thought that you could only find the Kernel if you have $a=b=c=0$? What am I missing here?
The kernel of an application $F$ is just the set of vectors $x$ that satisfies $F(x)=0$. Here, you have $$ F : Y=\begin{pmatrix} y_1\\ y_2 \\ y_3 \end{pmatrix} \mapsto \begin {pmatrix} y_2-y_3\\ 0 \\ 0 \end{pmatrix} $$ Hence, $Y \in \text{Kern}\left(F\right)$ if and only if $$ F\left(Y\right)=0 \Leftrightarrow y_2=y_3 $$ Hence, the kernel of $F$ is $$ \text{Kern}\left(F\right)=\begin{pmatrix} y_1\\ y_2 \\ y_2 \end{pmatrix} $$ Meaning that $(1,18,18)$ has a null image by $F$ but not $(1,2,3)$.