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Let $M$ be a set of matrices of integers. Let $R$ be the relation on $M$ defined as follows: For any two $m_1, m_2 \in M, (m_1, m_2) \in R$ iff the matrix multiplication $m_1 \cdot m_2$ is defined.

Part (a) - Given any element $m_1 \in M$, suppose that $(m_1, m_1) \in R$. What does that tell you about the shape of $m_1$?

For this part I'm pretty confident that the shape is a square.

Part (b) - Suppose that the elements of $M$ are chosen such that $R$ is reflexive. Explain why it must be true that $R$ is an equivalence relation.

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From part (a), you know all the elements of $M$ are square matrices. You just need to show that the relation is therefore symmetric and transitive.

Adding more detail:

We know all elements of $M$ are square matrices.

Symmetry: If $(m_1, m_2) \in R$, then the matrix multiplication $m_1 m_2$ is defined. Therefore both matrices must be the same size ($n \times n$ for some $n$). What can you conclude about $(m_2, m_1)$?

Transitivity: If $(m_1, m_2) \in R$, then by the same argument, $m_1$ and $m_2$ must be the same size ($n \times n$ for some $n$). Similarly if $(m_2, m_3) \in R$, then $m_2$ and $m_3$ are the same size. What can you say about $(m_1, m_3)$?

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  • $\begingroup$ Since it's reflexive we can say it is $(a,a) \in R$. Symmetry is $(b,a) ~ when~ (a,b) \in R$, but since its $(a,a) \in R$ that also makes it symmetric and the fact its a square. Then for Transitivity since its for any two $m_1, m_2$ we can say that there is a element in between them $m_{1.5}$ that we can say which they are all related to each other? Any nudges into the right direction would be awesome $\endgroup$
    – Helpless
    Jul 29, 2014 at 2:33
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    $\begingroup$ @Helpless I don't understand your comment at all $\endgroup$
    – angryavian
    Jul 29, 2014 at 2:37
  • $\begingroup$ @Helpless see my edits above $\endgroup$
    – angryavian
    Jul 29, 2014 at 2:53
  • $\begingroup$ Thank you so much! That is kind of what I was thinking about I just didn't know how start it and word it properly. $\endgroup$
    – Helpless
    Jul 29, 2014 at 3:29

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