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Show that set of $2 \times 2$ non-zero matrices over rationals does not form a ring under matrix addition and multiplication.

I am not able to see why this set won't be a ring. $(R,+)$ forms an abelian group. Associative and distributive property holds in $R$. So $(R,+,\times)$ should be a ring. Am i missing something ?

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  • $\begingroup$ non-zero <- there $\endgroup$ – Daniel Fischer Sep 16 '13 at 18:20
  • $\begingroup$ @DanielFischer Since that gave the OP his/her eureka moment, it's worth posting as an answer. Please ping me if you do put it up! $\endgroup$ – rschwieb Sep 17 '13 at 16:45
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    $\begingroup$ For people reading this: after reading Daniel's comment, the OP posted a comment like "oh of course now I see!" which was consequently deleted, probably because it contained profanity. I would like to imagine that Archimedes also added such things to his "eureka!" $\endgroup$ – rschwieb Sep 17 '13 at 17:57
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You missed a word in the problem statement:

Show that set of $2\times 2$ non-zero matrices over rationals does not form a ring under matrix addition and multiplication.

That makes the difference, the set contains no additive identity (and isn't even closed under addition or multiplication).

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  • $\begingroup$ if instead of "non-zero", the question says "rational" will it form a ring then ? $\endgroup$ – Aman Mittal Sep 19 '13 at 12:10
  • $\begingroup$ Yes, the set of all $n\times n$ matrices with entries in a ring, with the operations of matrix addition and matrix multiplication forms a ring. To get something that is not a ring, you must remove some matrices so that at least one of the axioms fails to hold. $\endgroup$ – Daniel Fischer Sep 19 '13 at 12:18
  • $\begingroup$ sorry, i again missed a word. the question wants me to prove that the set of all $2x2$ non-singular matrices over rationals is not a ring. I cant see why that will not be a ring. It should even have inverse. $\endgroup$ – Aman Mittal Sep 19 '13 at 12:21
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    $\begingroup$ The zero matrix is not non-singular. The sum of two non-singular matrices need not be non-singular (consider $A + (-A)$). $\endgroup$ – Daniel Fischer Sep 19 '13 at 12:24
  • $\begingroup$ oh !! Thank you very much for that. I was missing this point. $\endgroup$ – Aman Mittal Sep 19 '13 at 12:25

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