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Let

$$A \sim B \iff \exists v \in \mathbb{R}^2 \setminus \left( \begin{array}{c} 0\\ 0\\ \end{array} \right) : \mathcal{L}_{A,v} \cap \mathcal{L}_{B,v} \neq \emptyset$$

be a relation on $\mathbb{R}^{2 \times 2}$. $\mathcal{L}_{A,b}$ is defined as $\{ x \in \mathbb{R^n}: Ax = b\}$.

Now check if $\left( \begin{array}{cc} 1 & 0\\ 0 & 0\\ \end{array} \right) $ and $\left( \begin{array}{cc} 0 & 0\\ 0 & 1\\ \end{array} \right) $ are in relation to each other.

Given all of this, my conclusion is that I need to check if $Ax = Bx$, right? Since both $x$ are the same, this means I just need to check if $A=B$?

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    $\begingroup$ No, you need to check if there exists some $x$ such that $Ax=Bx\neq 0$. $\endgroup$ – Captain Lama May 14 at 10:15
  • $\begingroup$ So essentially $(A-B)x = 0$, but this results in the same statement $A=B$, doesn't it? $\endgroup$ – Max May 14 at 10:23
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    $\begingroup$ Not at all, I don't know why you would think that. A matrix $M$ can satisfy $Mx=0$ for some $x$ without being the zero matrix. $\endgroup$ – Captain Lama May 14 at 10:24

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