# Simple sub-space question given solution vector $(1,2,3)$ , space($M_{3x3}(\Bbb R)$) and a system $Ax=0$

The set of matrices $$A \in M_{3x3}(\Bbb R)$$ such that the vector $$(1,2,3)$$ is the solution for the system $$Ax=0$$ is a sub space of $$M_{3x3}(\Bbb R)$$

The book says that the statement is true and what I did was

$$Ax=0$$ and we have A= $$\left(\begin{matrix} a_{1,1} & a_{1,2} & a_{1,3} \\ a_{2,1} & a_{2,2} & a_{2,3} \\ a_{3,1} & a_{3,2} & a_{3,3} \\ \end{matrix} \right)$$

so we get $$A \cdot x$$= $$\left(\begin{matrix} a_{1,1} & a_{1,2} & a_{1,3} \\ a_{2,1} & a_{2,2} & a_{2,3} \\ a_{3,1} & a_{3,2} & a_{3,3} \\ \end{matrix} \right)$$ $$\cdot$$ $$\left(\begin{matrix} 1 \\ 2 \\ 3 \\ \end{matrix} \right)$$ $$=$$ $$\left(\begin{matrix} a_{1,1} + 2a_{1,2} + 3a_{1,3} \\ a_{2,1} + 2a_{2,2} + 3a_{2,3} \\ a_{3,1} + 2a_{3,2} + 3a_{3,3} \\ \end{matrix} \right)$$ $$=$$ $$\left(\begin{matrix} 0 \\ 0 \\ 0 \\ \end{matrix} \right)$$

then if we sort it as a system we get $$\begin{cases} a_{1,1} + 2a_{1,2} + 3a_{1,3}=0 \\ a_{2,1} + 2a_{2,2} + 3a_{2,3}=0 \\ a_{3,1} + 2a_{3,2} + 3a_{3,3}=0 \end{cases}$$

put it in a matrix again and we get $$\left(\begin{matrix} 1 & 2 & 3 \\ 1 & 2 & 3 \\ 1 & 2 & 3 \\ \end{matrix} \right)$$ $$=M$$

so lastly we get $$M \subseteq M_{3x3}(\Bbb R)$$

is this way correct ? if yes is there another way by using simple methods? thank you

• Note that the matrix you have is not correct as $\begin{bmatrix}1 & 2&3\\1 &2&3\\1&2&3\end{bmatrix}\begin{bmatrix}1\\2\\3\end{bmatrix}\neq0$, you should have that the rows look like $(1,1-1)$ or multiples of them, and the rows can be different Jan 7, 2022 at 23:28
• How can $M\subseteq M_{3\times3}(\Bbb R)$? After all, $M$ is a matrix, not a set of matrices. Jan 7, 2022 at 23:29

Let$$S=\{A\in M_{3\times3}(\Bbb R)\mid A.(1,2,3)=0\}.$$Asserting that $$M$$ is a subspace of $$M_{3\times3}(\Bbb R)$$ means two thins: that the sum of the elements of $$S$$ belongs to $$S$$ and thay if $$\lambda\in\Bbb R$$ and if $$A\in S$$, then $$\lambda A\in S$$. But:
• If $$A,B\in S$$, then$$(A+B).(1,2,3)=A.(1,2,3)+B.(1,2,3)=0+0=0,$$and therefore $$A+B\in S$$.
• If $$A\in S$$ and $$\lambda\in\Bbb R$$, then$$(\lambda A)(1,2,3)=\lambda\bigl(A.(1,2,3)\bigr)=\lambda.0=0,$$and therefore $$\lambda A\in S$$.