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

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?

• This is precisely what you do, and it's totally correct from what I know. If you want to look at all the vectors that are in the kernel, you can solve the equations $Av=0$ with $v=(a,b,c)^T$ and 0 the 0-vector. You can see the vector that spans the entire $\ker V$ and every vector of that form (with the $a,b,c$) will be in the kernel. Edit: note that you have to be careful with your notation; $v=(2,2,2)$ cannot be multiplied with that matrix, cause you need a column vector.
– Marc
May 23, 2018 at 20:49

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.
• @SkiMask Typically I will write a column vector vertically but you wrote it horizontally. To correct it you can write $\vec v=(2 \ , \ 2 \ , \ 2)^T$ where $T$ is the transpose which will be a column vector. May 24, 2018 at 1:40
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 $\mathbb{A}$: $$\mathrm{Ker}A=\{x\in R^n|\mathbb{A}x=0\}.$$