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We had a quiz earlier about binary systems. One problem got me thinking for the whole day.

Let $S$ be a subset of $M_{2}(\mathbb{R})$ consisting of all matrices of the form $\Bigg[\begin{matrix} a & a \\ a & a \\ \end{matrix}\Bigg]$, where $a\neq0$.

  1. Show that $S$ is a commutative binary structure under matrix multiplication.

  2. Does it have an identity element? If yes, find it.

  3. Given $A\in S$, does $A^{-1}$ exist in $S$?

I'm mostly interested in the third item, so I'll skip with the first one.

Define $\ast$ to be matrix multiplication. In algebraic structures, we know that an identity element exists if there exists a unique $e$ such that $a\ast e=a$. Since the identity matrix should be an element of $S$, then it should also be in the form of $\Bigg[\begin{matrix} e & e \\ e & e \\ \end{matrix}\Bigg]$. Hence, $$\Bigg[\begin{matrix} a & a \\ a & a \\ \end{matrix}\Bigg]\Bigg[\begin{matrix} e & e \\ e & e \\ \end{matrix}\Bigg]=\Bigg[\begin{matrix} a & a \\ a & a \\ \end{matrix}\Bigg]$$ So we have $2ae=a$. Therefore, the identity element is $\Bigg[\begin{matrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \\ \end{matrix}\Bigg]$. For the case of inverse, we can show that the inverse exists if $a\ast a'=e$. Therefore, $$\Bigg[\begin{matrix} a & a \\ a & a \\ \end{matrix}\Bigg]\Bigg[\begin{matrix} a' & a' \\ a' & a' \\ \end{matrix}\Bigg]=\Bigg[\begin{matrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \\ \end{matrix}\Bigg]$$ So we have $2aa'=\frac{1}{2}$. Hence, its inverse is $\Bigg[\begin{matrix} \frac{1}{4a} & \frac{1}{4a} \\ \frac{1}{4a} & \frac{1}{4a} \\ \end{matrix}\Bigg]$.

I'm not sure if this is right or wrong. I'm convinced that this is wrong since we can show that the determinant is equal to zero $(a^{2}-a^{2}=0)$. Also, I don't know if this is relevant, but we've been taught the identity matrix for multiplication is always $\Bigg[\begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix}\Bigg]$, but since it's not an element of $S$, I don't really know what to do with it.

Here are some of my questions:

  1. Was it okay for me to assume that $\ast$ is defined to be the matrix multiplication? I think it was wrong, but I also don't know where to go with this problem if it's an arbitrary operation.

  2. What does the determinant say about a matrix? Why can't it be equal to zero for a matrix to have an inverse?

  3. If wrong, what is the correct way of proving or disproving its existence?

Please don't use concepts from linear algebra. Just limit it to basic group theory and basic matrix concepts. Any answer would be much appreciated.

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  • $\begingroup$ Note that your operation is essentially $(a,b) \mapsto 2ab$ once you record only the northwest entry. Now use this to get all your assertions. $\endgroup$
    – Randall
    Aug 25, 2017 at 1:35

2 Answers 2

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  • "Was it okay for me to assume that ∗ is defined to be the matrix multiplication?"

Yes, because you were told to "how that S is a commutative binary structure under matrix multiplication"

  • "What does the determinant say about a matrix?" (and related questions)

The identity of a binary operation is an element that doesn't change other elements, so $1\star a=a\star 1=a$ for all $a$.

If you consider the set $M$ of all 2x2 matrices, there's only one matrix $I$ for which $AI=IA=A$ for all $A$.

Now, though, you're looking at $S$. Your matrix $E=\Bigg[\begin{matrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \\ \end{matrix}\Bigg]$ satisfies $EA=AE=A$ for all $A\in S$, so it's an identity within $S$. It's clearly not an identity within $M$, since there are plenty of matrices in $M$ that don't have $EA=AE=A$.

So, if you have a ring structure, there's nothing stopping proper subsets of it from having completely different identity elements. In fact, any time you have an element A of a ring with $A^2=A$, you can find a set of elements of the ring for which $A$ is an identity.

The identity of $M$ is $I$. The identity of $S$ is $E$.

You've worked out inverses in $S$ correctly - every element (except $O$) has an inverse, and you have the right formula for them. Yes, they all have determinant zero, and still have inverses in $S$. They don't have inverses in $M$, but they do in $S$. The determinant is a nice tool for figuring out which matrices have inverses in $M$, but it's no use in $S$, you'll need a different criterion. Maybe $\det_S\Bigg[\begin{matrix} a & a \\ a & a \\ \end{matrix}\Bigg]=2a$ will do.

Actually, though, you don't need determinants at all for this question. There's a simple rule for testing invertibility in $S$, namely, $a\neq0$, and the traditional determinant $ad-bc$ is not relevant since it's talking about $M$.

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Direct answers to your questions

Regarding 1: The first question says "show that $S$ is a commutative binary structure under matrix multiplication." It is therefore extremely likely that, for the rest of the question, the binary operation is still supposed to be matrix multiplicaiton.

Regarding 2: The inverse of a matrix in the linear-algebra sense is the inverse of a matrix within the binary structure $M_2(\Bbb R)$ under matrix multiplication. In particular, the "inverse" of a matrix $A$ in that context is a matrix $B$ such that $AB = BA = I$ (where $I = \left( \begin{smallmatrix} 1&0\\0&1 \end{smallmatrix} \right)$). Notably, $I$ is the unique identity element within that structure.

Our question, however, is about the smaller binary structure $S$. You have correctly identified the identity element $E = \left( \begin{smallmatrix} 1/2&1/2\\1/2&1/2 \end{smallmatrix} \right)$ within this structure. By definition, the inverse of an element $A \in S$ with respect to the binary structure on $S$ is the element $B$ such that $AB = BA = E$. You have found the correct formula for this inverse.

Note, however, that the elements $A \in S$ do not have an inverse with respect to the broader structure on $M_2(\Bbb R)$. In particular, there is no matrix $B \in M_2(\Bbb R)$ such that $AB = BA = I$. In general, any matrix $A$ whose determinant is zero cannot have an inverse with respect to the multiplicative binary structure on $M_n(\Bbb R)$.

So long story short: although you are convinced it is wrong, you were actually right.

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