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Let $M_2(\mathbb{R})$ denote the set of all $2 \times 2$ matrices obviously with all Real entries.

let $R =$ $$ \begin{bmatrix} a & b \\ -b & a \\ \end{bmatrix} $$

Prove that $R$ is subring of $M_2(\mathbb{R})$

By definition a subrinng S of a ring R is a subgroup of R under the addition it inherits from 𝑅 which is closed under the inherited multiplication and has the same multiplicative identity as 𝑅. In order to show $R$ is a subgroup I need to show the following:

1) R is closed under addition and multiplication

2) R is associative

3) R has an identity element

4) Every element has an inverse

Number 1)

Well $R+R$ yields:

$$ \begin{bmatrix} 2a & 2b \\ -2b & 2a \\ \end{bmatrix} $$

and

$R \times R$ yields:

$$ \begin{bmatrix} a^2 - b^2 & 2ab \\ -2ab & b^2-a^2 \\ \end{bmatrix} $$

Since $2a,2b,-2b \in \mathbb{R}$ and since $a^2-b^2,2ab,-2ab,b^2-a^2 \in \mathbb{R}$ It follows that R is closed under addition and multiplication

Number 2: I am not sure how to say that it is associative. I should be able to imply this from the fact that it is in the set of all $2 \times 2$ matrices

Number 3: $R$ has the identity element $I_2$ which is denoted as:

$$ \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} $$

Number 4: By definition the inverse of a $2 \times 2$ matrix is:

$\frac{1}{ad-bc} \times$

$$ \begin{bmatrix} d & -b \\ -c & a \\ \end{bmatrix} $$ So in the case of $R+R$ the inverse is $\frac{1}{4a^2+4b^2} \times$:

$$ \begin{bmatrix} 2a & -2b \\ 2b & 2a \\ \end{bmatrix} $$ and in the case of $R \times R$ the inverse is $\frac{1}{((a^2-b^2)(b^2-a^2))-(4a^2b^2)} \times$

$$ \begin{bmatrix} b^2-a^2 & -2ab \\ 2ab & a^2-b^2 \\ \end{bmatrix} $$

Does this work properly show that $R$ is a subgroup of $M_2(\mathbb{R})$

Clearly $I_2$ which is the multiplicative dentity of $R$ is also the same multiplicative identity of $M_2(\mathbb{R})$ which is mentioned in the definition.

Does adding this last part in with the rest of the work show that $R$ is infact a subring?

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You do way too much work. By the subring criterion, it is both sufficient and necessary to check $$1 \in R$$ $$x,y \in R \implies x-y \in R$$ $$x,y \in R \implies xy \in R$$

all of which follow after an easy computation.

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I'm afraid there's a fundamental misunderstanding in your work, starting with your saying

let $R=\begin{bmatrix}a&b\\-b&a\end{bmatrix}$.

This says that $R$ is just a single matrix, but that's not what it's supposed to be here. If you read the question that was given to you more carefully, you should see that $R$ is the set of all matrices of this form: $$R=\left\{\begin{bmatrix}a&b\\-b&a\end{bmatrix}:a,b\in\mathbb{R}\right\}\subseteq M_2(\mathbb{R}).$$

Because of this misunderstanding, your work on proving the required conditions is not correct.

Let's start with the first one. Yes, you need to show that for any two matrices from $R$ their sum is again in $R$. In other words: show that if $X,Y\in R$, then $X+Y\in R$. That's not what you did; you only took one matrix from $R$ and multiplied it by $2$.

For a correct proof of this property, you need to start with two generic matrices from $R$, say $X=\begin{bmatrix}a&b\\-b&a\end{bmatrix}$ and $Y=\begin{bmatrix}c&d\\-d&c\end{bmatrix}$. Then their sum is $$X+Y=\begin{bmatrix}a&b\\-b&a\end{bmatrix}+\begin{bmatrix}c&d\\-d&c\end{bmatrix}=\begin{bmatrix}a+c&b+d\\-b-d&a+c\end{bmatrix}=\begin{bmatrix}a+c&b+d\\-(b+d)&a+c\end{bmatrix}\in R,$$ because it has the required form of being a member of $R$ (the two entries on the main diagonal are equal, and the two entries on the other diagonal are negatives of each other).


As others have pointed out, there's a shorter list of properties one needs to verify in order to prove that a subset of a ring is a subring. But first of all, the correct understanding of the question is crucial. I hope this explanation helps. Now using this approach, you can work on the remaining properties. Best of luck!

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  • $\begingroup$ Thank you for the clarification. You made it clear what the issue was. $\endgroup$ – BlackKnightRider Mar 30 '20 at 18:42
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A non-empty subset $S$ of a ring $R$ is a subring iff $x - y, xy \in S$ for all $x, y \in S$. You have shown that you have a non-empty subset. What you still have to show is that the difference and the product of two such matrices is in the subset. Therefore, you have to consider two arbitrary matrices from $R$, take for example

$$ \begin{pmatrix} a & b \\ b & - a \end{pmatrix}, \qquad \begin{pmatrix} c & d \\ d & - c \end{pmatrix}.$$

Note that your step 4 is not needed since you don't have to show the invertibility with respect to the multiplication (but with respect to addition).

Remark: The associativity of addition and multiplication in $R$ is clear since this is in general the case for real matrices.

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  • $\begingroup$ So all I have to show is that $R \times R$ is in the subset as I did and then I show the difference between what? $\endgroup$ – BlackKnightRider Mar 30 '20 at 18:27
  • $\begingroup$ Yes, you have shown that the product of two elements of $R$ is in $R$. What you still have to show is that if you take two matrices from $R$, their difference is also in $R$. $\endgroup$ – Jan Mar 30 '20 at 18:29
  • $\begingroup$ so I can use $a_1$ and $a_2$ and $b_1$ and $b_2$? $\endgroup$ – BlackKnightRider Mar 30 '20 at 18:31
  • $\begingroup$ Yes, this is possible. And now I see: You also have to do this for the product, your computation does not work at the moment. $\endgroup$ – Jan Mar 30 '20 at 18:31
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    $\begingroup$ Call your matrix at the top $A$. You have shown that $A + A = 2A \in R$ and $A \cdot A = A^2 \in A$. But what you have to show is that $A + B, AB \in R$ for $A, B \in R$. The second matrix has to differ from $A$. $\endgroup$ – Jan Mar 30 '20 at 18:40

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