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Determine whether the indicated set forms a ring under the indicated operations. $S = \{A ∈ M(2,\mathbb R) \mid \det A = 0\}$, under matrix addition and multiplication.

I'm not entirely sure about this but when I was going through the ring axioms, I noticed that if you take $A^ {-1} = \frac {1}{0} $ $\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$, which is not possible. Since that axiom fails, this isn't a ring under addition and multiplication. Similarly, there is no $\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$ in this group. There isn't really a unique identity element per se but any $A,B \in S$ under $det (A + B) = det A + det B = det A + 0 = det A = 0$. So can I say that any element in the set acts as an identity?

My other thought was that the definition of the inverse for a group in general is $∀a∈G:∃b∈G: a∘b=e=b∘a$. But for any $ A, B \in S$, $det (A + B) =0$. So perhaps the inverse I mentioned earlier could be any other element in $S$ and we can completely disregard taking the inverse of $A$ because anything can act as the inverse.

I suppose I am really confused on how to interpret the identity and inverse for this particular problem. Any help resolving this would be appreciated.

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  • $\begingroup$ Have you confirmed $\det(A+B)=\det(A)+\det(B)$ for any concrete matrices? $\endgroup$ – Arthur Apr 12 '16 at 6:24
  • $\begingroup$ Yes. I showed that closure, commutativity, and associativity work for this set. Ok. So I'm a little confused. I get that the 0 matrix is in S. So we have an identity element. Now I know we can't take inverses of elements but I guess taking the negative of each matrix would yield the identity. So then in terms of ring, we have an inverse. But what is the other axiom that fails?@Arthur $\endgroup$ – TfwBear Apr 12 '16 at 6:26
  • $\begingroup$ Let me be more specific. I have done all the axioms for addition but not yet multiplication. $\endgroup$ – TfwBear Apr 12 '16 at 6:27
  • $\begingroup$ Multiplicative inverses aren't required in the definition of a ring. $\endgroup$ – manthanomen Apr 12 '16 at 6:27
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Remember the checklist for rings:

  • Is addition associative, that is, $(a+b)+c=a+(b+c)$?
  • Is addition comutative, that is, $a+b = b + a$?
  • Is there a zero for the addition, that is, some element $0$ so that $a + 0 = a$ for all $a$?
  • Does each element have an additive inverse, that is, for all $a$ is there some $b = -a$ with $a + b = 0$?
  • Is multiplication associative?
  • Is there a multiplicative identitity, that is, some element $1$ so that $a * 1 = 1 * a = a$ for all $a$?
  • Is multiplication distributive with respect to addition (left and right)?

It it easy to see the answer to all of these is "Yes" in your case, with $0$ being the matrix whose entries are all $0$ and $1$ being the identity matrix. Notice in particular that it is not necessary that any element have a multiplicative inverse (that is $a^{-1}$ need not exist).

So, everything points out to our set actually being a ring, right? We must not forget, however, that our set must be closed under addition and multiplication, meaning that these operations acting on elements of our set should always produce elements that still belong to it.

Now, you know $\det(A\cdot B) = \det(A)\cdot\det(B)$, so it is closed under multiplication. But what about addition? Can you think of a pair of matrices with determinant $0$ that add up to a matrix nonzero determinant?

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  • $\begingroup$ Ok I see. So there might be some matrices that individually have det 0 but under addition not the case. $\endgroup$ – TfwBear Apr 12 '16 at 6:32
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Hint: calculate the determinant of $$ \begin{bmatrix}1&0\\0&0\end{bmatrix} + \begin{bmatrix}0&0\\0&1\end{bmatrix} $$

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