I am trying to determine whether or not the set of all n x n diagonal matrices under matrix multiplication is a group. I can show that the set is closed and associative under the operation, but I am confused on how to show that every element of the set has an inverse in the set. I understand that since the determinant of any element of the set is nonzero each element has an inverse, but why must that inverse belong to the set?

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you seem to have neglected the "the determinant of the matrix is non-zero" from the beginning of the question. If you are considering the collection of diagonal matrices, then it is not a group. But the collection of invertible diagonal (i.e., those diagonal matrices with non-zero determinant), then it is indeed a group. To see that this collection is closed under inverses, take an arbitrary diagonal matrix $diag(d_1,\ldots, d_n)$ with non-zero determinant. What does that tell you about each $d_i$? Can you now explicitly describe the inverse matrix and see that it too is diagonal?

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  • $\begingroup$ The definition of a diagonal matrix I was given was that each diagonal entry must be nonzero and every other entry must be zero. Is this incorrect? Is it correct to say that under this definition every diagonal matrix is invertible? $\endgroup$ – user155242 Sep 24 '14 at 1:02
  • $\begingroup$ this is not the standard definition. A diagonal matrix (the way is commonly defined) is a square matrix all of whose off-diagonal entries are $0$. According to the non-standard definition you quote it is indeed the case that every diagonal matrix is invertible. Needless to say, you should adhere to the standard uses! $\endgroup$ – Ittay Weiss Sep 24 '14 at 1:19

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