Does vector $v$ lie in the Kernel of $A$ (Matrix)?

Question

I want to check if vector $v$ is in the Kernel of $A$ (Matrix) and to do that I used the definition of Kernel, $$Kern(V):=\left\{{{\vec v \in {R^3}} \ | L(\vec v)=\vec 0} \ \right\}$$ If $\vec v=(2 \ , \ 2 \ , \ 2)$ and $A=\begin{pmatrix}1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5\end{pmatrix}$, then $A\vec v=\begin{pmatrix}12 \\ 18 \\ 24 \end{pmatrix}≠\begin{pmatrix}0 \\ 0 \\ 0 \end{pmatrix}$, therefore $\vec v$ doesn't lie in the Kernel of $A$. Is this correct since I'm using the definition of the Kernel. Is my answer correct?

Answer 1

Just a few points on your answer:

1). Your definition of the kernel is incorrect in the sense that you have defined the kernel of a matrix $V$, but a matrix $L$ is used as well. What you want is $$ \mathrm{ker}(A) = \{v \in \mathbb{R}^3 \ : \ Av=0\}. $$

2) The way you have written $v$ means that $Av$ is undefined, you cannot multiply a 3x3 and a 1x3 matrix. You should write $v$ using column notation.

Otherwise, your answer is correct.

Answer 2

A matrix $A\in R^{n\times m}$ can be seen as a map, denoted by $\mathbb{A}$, from $R^n$ to $R^m$. The Kernel of $A$ is defined by the Kernel of the map $\mathbb{A}$: $$\mathrm{Ker}A=\{x\in R^n|\mathbb{A}x=0\}.$$