# Show that ~ creates a partition of $M_2(\mathbb{R})$

Let $M_2 (\mathbb{R})$ be the set of 2x2 matrices over $\mathbb{R}$: $$M_2 (\mathbb{R}) = \biggl\{ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \; \biggm| \; \text{with } a,b,c,d \in \mathbb{R} \biggr\}.$$ For $M_1, M_2 \in M_2(\mathbb{R})$, we say that $M_1 \sim M_2$ if $\det(M_1) = \det(M_2)$, where $$\det \begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad-bc.$$

• Show that $\sim$ creates a partition of $M_2(\mathbb{R})$.
• What are the representative elements of each partition?
• Are there countably or uncountably many distinct equivalence classes?

I am stuck and I could really use some help. I know that in order to have a partition, the union of all the partitions $X_i$ should be $X$ and $X_i \cap X_j = \varnothing$ for $i \ne j$.

Hints:

• It suffices to show that $\sim$ is an equivalence relation on $M_2(\mathbb R)$. So simply show that it's reflexive, symmetric, and transitive. This follows immediately, since $\sim$ is defined in terms of $=$.

• Try some examples. Pick a matrix at random, and find its equivalence class. Then try going backwards: pick a determinant at random, then try to find a matrix that has that as a determinant.

• From the above bullet, try to show that each $x \in \mathbb R$ corresponds to a distinct equivalence class $[M]_\sim$.

Whenever we have a function $f\colon X\to Y$, we can define a relation $\sim_f$ over $X$ by declaring $$a\sim_f b\qquad\text{if and only if}\qquad f(a)=f(b)$$ This is readily seen to be an equivalence relation:

• for all $a\in X$, $a\sim_f a$: indeed $f(a)=f(a)$
• for all $a,b\in X$, if $a\sim_f b$, then $b\sim_f a$: indeed $f(a)=f(b)$ implies $f(b)=f(a)$
• for all $a,b,c\in X$, if $a\sim_f b$ and $b\sim_f c$, then $a\sim_f c$: indeed $f(a)=f(b)$ and $f(b)=f(c)$ implies $f(a)=f(c)$

The equivalence class of $a\in X$ is $a/{\sim_f}=\{x\in X:f(x)=f(a)\}$.

Your case is with $X=M_2(\mathbb{R})$, $Y=\mathbb{R}$ and $f$ is the determinant function.

“The” representative is not well defined. What you need is, given finding one representative, but there is no “canonical” choice. Since any real number can be the determinant of a suitable matrix, given $r\in\mathbb{R}$ you just need a matrix $A$ such that $\det A=r$.

Note also that, in the general case, one can define a map $$\bar{f}\colon X/{\sim_f}\to Y$$ by declaring that $\bar{f}(a/{\sim_f})=f(a)$. This map is injective and has the same image as $f$. So, in your particular case, the set of equivalence classes is in 1-1 correspondence with $\mathbb{R}$.

Compute $$\det\pmatrix{a&0\\0&1}$$ What do you notice?
The partition property is inherent because $\det$ is a function and $x \sim_f y :\Leftrightarrow f(x) = f(y)$ is an equivalence relation for any function $f$.