Algebra: show that the n x n square matrices such that in each row and in each column of them...?

Question

... there is one element equal to 1 and the other elements are zeros from a group under matrix multiplication?

Answer 1

Let G be the set of matrices that you described above. You need to show that G satisfies the group axioms.

First, you must show that G is closed under the multiplication operation. In other words, if you take any matrix from G, and multiply it by any other matrix in G, you will get a matrix that is a member of G. This is not too difficult to show, from the rules of matrix multiplication.

Second, you must show that the multiplication operation of G is associative. That is: (a * b) * c = a * (b * c) for any a, b, and c that are members of G. Well that's easy, because we already know that matrix multiplication is associative.

Third, you must show that G contains an identity element I, such that I * a = a * I = a for any a in G. That's easy too, because the identity matrix is a member of G. (Identity matrix is the one with 1's on the main diagonal, and 0's elsewhere.)

Fourth, you must show that every element of G has an inverse element that is also in G. That is, for any 'a' there is a 'b' such that:
a * b = b * a = I
You can show this from the rules of matrix multiplication. To begin, we know that none of the elements of G is a singular matrix. (All of them have determinant = 1 or -1) That means that each element of G has an inverse. You know that the inverse exists, so all that's left is to show that the inverse is also a member of G.
The members of G have a special property: For any matrix x which is in G, inverse(x) = transpose(x). It is easy to see that whenever x is a member of G, then transpose(x) is also in G. (The transpose operation will not alter the fact that there is only a single 1 in each row and in each column.) Therefore, inverse(x) is a member of G.

That should do it. You will have to fill in some details which I skimmed over in step 1 and step 4. Good luck. Hope that this helps you.

Answer 2

i do no longer understand the names as sturdy as i'm no longer English :-/ besides the undeniable fact that I doubt such lemmas can actually have any names. maximum of them are extremely user-friendly with obvious proofs. A: The inverse matrix of a created from matrices is the fellow made from their inverses in opposite order. as a effect, (C(D^T)B)^-a million = B^-a million (D^T)^-a million C^-a million Associativity of matrix multiplication, that you quietly used in the question, on the fringe of the definition of what an inverse matrix and effect of multiplication with id matrix provides B (B^-a million (D^T)^-a million C^-a million) C (D^T) B = (B B^-a million) ((D^T)^-a million (C^-a million C) (D^T)) B = I ((D^T)^-a million I D^T) B = ((D^T)^-a million D^T) B = I B = B. We did not choose right here that C is inverse matrix of B. B: B C = I (B C)^T = I C^T B^T = I^T = I B is skew-symmetric => B^T = -B C^T (-B) = I Linearity: - C^T B = I this promises that -C^T is the inverse matrix of B, it truly is given uniquely. although, all of us understand from above that C is a similar. for this reason, C = -C^T C^T = -C through this that C is skew-symmetric. Edit: in a), we used the undeniable reality that (D^T)^-a million exists, or, that D^T is invertible. remember with a view to operate to the reasoning that there might want to properly be yet another theorem that D is invertible <=> D^T is invertible (and it holds that (D^T)^-a million = (D^-a million)^T, which we do not opt for above). word to fretty: A matrix A which satisfies A^T = A^-a million isn't referred to as skew-symmetric yet antiorthogonal. this can properly be a touch uncommon resources.