How to find a basis for $W = \{A \in \mathbb{M}^{\mathbb{R}}_{3x3} \mid AB = 0\}$ $B = \begin{bmatrix}
    1 & 2  & 1 \\
    1 & 3  & 1 \\
    1 & 4  & 1
\end{bmatrix}$ and I need to find a basis for $W = \{A \in \mathbb{M}^{\mathbb{R}}_{3x3} \mid AB = 0\}$ .
I know that $AB = A\cdot\begin{bmatrix} 1\\1\\1 \end{bmatrix} \mid A\cdot\begin{bmatrix} 2\\3\\4 \end{bmatrix} \mid A\cdot\begin{bmatrix} 1\\1\\1 \end{bmatrix} = 0 \mid 0 \mid 0$
Then I can conclude that (assume $A_1,...,A_n$ are columns of $A$):
1) $A_1 + A_2 + A_3 = 0$
2) $2A_1 + 3A_2 +4 A_3 = 0$
Meaning: $A_1 + 2A_2 +3A_3 = 0$
But now I got stuck... How should I continue from here?


Then I can conclude that (assume $A_1,...,A_n$ are columns of $A$):
  1) $A_1 + A_2 + A_3 = 0$
  2) $2A_1 + 3A_2 +4 A_3 = 0$

Close! Instead of rows they should be columns because matrix multiplication would take the dot products of the row vectors of the first matrix with the column vectors of the second.
By elementary row operations, $(-2 eq. 1 + eq. 2)$ $A_2+2A_3=0 \Rightarrow A_2=-2A_3$
Substituting this relation into $eq. 1$ gives us $A_1-A_3=0 \Rightarrow A_1=A_3$.
So, we can create a matrix that respects the previous conditions:
$A =
\begin{bmatrix}
 a&b  &c \\ 
 -2a&-2b  &-2c \\ 
 a&b  &c 
\end{bmatrix}
$ and we want to find such $a,b,c$ that $AB=0$
$\begin{bmatrix}
 a&b  &c \\ 
 -2a&-2b  &-2c \\ 
 a&b  &c 
\end{bmatrix}\cdot \begin{bmatrix}
    1 & 2  & 1 \\
    1 & 3  & 1 \\
    1 & 4  & 1
\end{bmatrix}=\begin{bmatrix}
a+b+c&2a+3b+4c&a+b+c\\
-2(a+b+c)&-2(2a+3b+4c)&-2(a+b+c)\\
a+b+c&2a+3b+4c&a+b+c
\end{bmatrix}$
See how the system we are trying to solve: $a+b+c=0, \quad\!\! 2a+3b+4c=0$ is similar to the one above with $A_1,A_2,A_3$. So, $-b=2a=2c$.
The original matrix $A$ can now be expressed as $A =
\begin{bmatrix}
 t&-2t  &t \\ 
 -2t&4t &-2t \\ 
 t&-2t  &t 
\end{bmatrix}
$ where $t\in \mathbb{R}$
To construct a basis, notice that the matrix function $A(t)$ is of single variable. Thus, any multiple of \begin{bmatrix}
 1&-2  &1 \\ 
 -2&4 &-2 \\ 
 1&-2  &1 
\end{bmatrix} forms a basis for $W$.
$$W=span\begin{bmatrix}
 1&-2  &1 \\ 
 -2&4 &-2 \\ 
 1&-2  &1 \\
\end{bmatrix}=t\begin{bmatrix}
 1&-2  &1 \\ 
 -2&4 &-2 \\ 
 1&-2  &1 \\
\end{bmatrix}{\Huge{|}}t\in \mathbb{R}$$
