# Check, if A ~ B have a relation

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 $$A=B$$?

• No, you need to check if there exists some $x$ such that $Ax=Bx\neq 0$. – Captain Lama May 14 at 10:15
• So essentially $(A-B)x = 0$, but this results in the same statement $A=B$, doesn't it? – Max May 14 at 10:23
• 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. – Captain Lama May 14 at 10:24