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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

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    $\begingroup$ 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 $\endgroup$ Jan 7, 2022 at 23:28
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    $\begingroup$ How can $M\subseteq M_{3\times3}(\Bbb R)$? After all, $M$ is a matrix, not a set of matrices. $\endgroup$ Jan 7, 2022 at 23:29

1 Answer 1

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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 $\lambda A\in S$.
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