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Another matrix question!

Let $$S=\{A \in M_2(\mathbb{R}):f(A)=0\}\text{ and }f\left(\begin{bmatrix}a&b\\c&d \end{bmatrix}\right)=b$$ Is S a group under matrix addition. Either prove that (S,+) is a group or show that one of the group axioms fails.

Group axioms:

(G1) the set G is closed under the operation $*$

(G2) The operation $*$ is associative - $(a*b)*c=a*(b*c)$

(G3) There exists an element $e \in G$ such that $a*e=a=e*a$ for all $a \in G$

(G4) for each $a \in G$ there exists an element $a^{-1}$ of G such that $a*a^{-1}=e=a^{-1}*a$

Here's what I've got so far:

(G1) let $A=\begin{bmatrix}a&0\\c&d\end{bmatrix}$ and $B=\begin{bmatrix}e&0\\g&h \end{bmatrix}$ where $A, B \in S$ then $A+B=\begin{bmatrix}a+e&0\\c+g&d+h \end{bmatrix}$ which is $\in S$ so it is closed. - (G1) is satisfied

(G2) using $A$ and $B$ as above and $C=\begin{bmatrix}j&0\\m&n \end{bmatrix},\\ (A+B)+C = \begin{bmatrix}a+e&0\\c+g&d+h \end{bmatrix}+\begin{bmatrix}j&0\\m&n \end{bmatrix} = \begin{bmatrix}a+e+j&0\\c+g+m&d+h+n\end{bmatrix}$

and $A+(B+C)=\begin{bmatrix}a&0\\c&d\end{bmatrix}+\begin{bmatrix}e+j&0\\g+m&h+n\end{bmatrix}=\begin{bmatrix}a+e+j&0\\c+g+m&d+h+n\end{bmatrix}$

so $(A+B)+C=A+(B+C)$ - (G2) is satisfied

(G3) let $e=\begin{bmatrix}0&0\\0&0\end{bmatrix}$ pretty easy to see that $A+e=A=e+A$ - (G3) satisfied

(G4) let $A^{-1}=\begin{bmatrix}-a&0\\-c&-d\end{bmatrix}$

$A+A^{-1}=\begin{bmatrix}a&0\\c&d\end{bmatrix}+\begin{bmatrix}-a&0\\-c&-d\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}=e$ and $A^{-1}+A=\begin{bmatrix}-a&0\\-c&-d\end{bmatrix}+\begin{bmatrix}a&0\\c&d\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}=e$ (G4) satisfied

So I have proved all the axioms (hopefully correctly?!) therefore we can say that (S,+) is a group.

I know that's a bit of reading but I would really appreciate your confirmation that this is right / where I went wrong if it isn't. The one I'm really worried about is G1, and I know then that if this is wrong and (S,+) is not closed then I don't need to worry about the rest. And also if there is another way to do it?

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    $\begingroup$ Something is missing from the problem. You seem to be assuming that the condition $f(A)=0$ means that the upper-right entry of $A$ is zero. Without knowing what $f$ is supposed to be, it is hard to evaluate this. $\endgroup$ – mweiss Jul 24 '14 at 23:29
  • $\begingroup$ sorry yup I've added in $f$ now - it was from a previous part of the question so obviously I assumed you all knew that as well :P duh $\endgroup$ – panderson Jul 24 '14 at 23:30
  • $\begingroup$ Yes, this looks fine; well done! $\endgroup$ – user61527 Jul 24 '14 at 23:30
  • $\begingroup$ $f$ is a linear map and $\ker f=S$ is a vector subspace $\endgroup$ – Hamou Jul 24 '14 at 23:32
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This is great! So here's some feedback on the presentation:

  1. Proof of (G1):

    • This reads backwards in my opinion: "let $A=\begin{bmatrix}a&0\\c&d\end{bmatrix}$ and $B=\begin{bmatrix}e&0\\g&h \end{bmatrix}$ where $A, B \in S$..." I suggest writing: "let $A, B \in S$. Then, by definition, $A=\begin{bmatrix}a&0\\c&d\end{bmatrix}$ and $B=\begin{bmatrix}e&0\\g&h \end{bmatrix}$ where $a,c,d,e,g,h \in \mathbb{R}$. ..."

    • "...which is $\in S$ so it is closed" could be extended to: "...which shows $f(A+B)=0$ so $A+B \in S$, and $S$ is closed under matrix addition."

  2. Again, I suggest writing "let $C \in S$, so $C=$...". Also " - (G2) satisfied" is a bit abbreviated.

  3. Nothing to add.

  4. "let $A^{-1}=\begin{bmatrix}-a&0\\-c&-d\end{bmatrix}$" This should be denoted $-A$ not $A^{-1}$. Additionally, the notation $-A$ and $A^{-1}$ somewhat presupposes that an inverse exists.


Further comment:

  • If we know $(M_2(\mathbb{R}),+)$ is a group, then we need only to check that $S$ is a subgroup. We could use a subgroup test, which is a bit quicker. We just check (a) that $S$ is non-empty, and (b) that if $A,B \in S$, then $A-B \in S$ too.
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