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Triviality of Ideals in M(n,R).

  1. Oct 31, 2014 #1

    WWGD

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    Hi All,
    I am trying to show that the only ideals in M (2, ## \mathbb R ##) , the ring of 2x2 matrices with Real entries are the trivial ones.

    I have a proof, but I am being kind of lazy rigorizing it. We know we cannot have any matrix in GL(n,##\mathbb R ##), because we can then get the identity and we end up with the whole ring. Basically then, we take any
    non-invertible matrix m and we show we can find matrices A,A' in M (2, ## \mathbb R ##) so that Am + Am' is invertible. Is there a way of tightening this?

    I thought of using the result that maybe either GL(n, ## \mathbb R ## ), or maybe
    M (2,##\mathbb R ##) acts transitively on the left on M (2, ## \mathbb R ##) by multiplication. Is this result true?

    Thanks.
     
  2. jcsd
  3. Oct 31, 2014 #2

    jbunniii

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    Certainly ##GL(2,\mathbb{R})## does not act transitively on ##M(2,\mathbb{R})##. For if ##A## is not invertible, and ##M## is invertible, then ##X = MA## is not invertible (if it were, then ##A = M^{-1}X## would also be invertible). Therefore, if ##A## is not invertible and ##B## is invertible, then ##A## and ##B## are not in the same orbit under this action.

    ##M(2,\mathbb{R})## is not a group with respect to multiplication, so you have to clarify what you mean by "##M(2,\mathbb{R})## acts on the left".

    I don't know of an especially clean proof; the elementary one is straightforward: if ##A \in M(2,\mathbb{R})## has a nonzero entry in row ##i##, column ##j##, then we can premultiply and postmultiply by appropriate elements of ##M(2,\mathbb{R})## to obtain a matrix which has the same nonzero entry at ##i,j## and which has zeros everywhere else. If this nonzero entry is, say, ##a##, then we can multiply (on the left or right) by ##a^{-1}I## to get a matrix with ##1## at ##i,j## and ##0## elsewhere. Then we can move this ##1## wherever we like by multiplying on the left by a row-swapping matrix or on the right by a column-swapping matrix. Therefore, ##\begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}## and ##\begin{pmatrix}0 & 0 \\ 0 & 1\end{pmatrix}## are in the ideal generated by ##A##, and as a result, so is ##I##. So the ideal contains a unit, and is therefore the whole ring.
     
    Last edited: Oct 31, 2014
  4. Oct 31, 2014 #3

    WWGD

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    Thanks, sorry, I meant "act" in an informal sense, meaning given any A,B there is C with AC=B , and I meant on M(n,R)- GL(n, R).
     
  5. Oct 31, 2014 #4

    jbunniii

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    If you were to make an argument using group actions, I think you would have to consider actions by multiplication on the left and on the right, because ##M(2,\mathbb{R})## does have nontrivial one-sided ideals. For example, the set of all matrices in ##M(2,\mathbb{R})## with zeros in the second column is a left ideal.
     
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