(We say B is an inverse of A.) Dear Teachers, Students and Parents, We are presenting here a New Concept of Education, Easy way of self-Study. If two matrices commute: AB=BA, then prove that they share at least one common eigenvector: there exists a vector which is both an eigenvector of A and B. Let A be m n, and B be p q. In this case by the first theorem about elementary matrices the matrix AB is obtained from B by adding one row multiplied by a number to another row. Prove that if A and B are diagonal matrices (of the same size), then. 3) For A to be invertible then A has to be non-singular. Prove that if A and B are diagonal matrices (of the same size), then AB = BA. AB = BA.. Getting Started: To prove that the matrices AB and BA are equal, you need to show that their corresponding entries are equal. Theorem 3 Given matrices A 2Rm l, B 2Rl p, and C 2Rp n, the following holds: A(BC) = (AB)C Proof: Since matrix-multiplication can be understood as a composition of functions, and since compositions of functions are associative, it follows that matrix-multiplication Hence both AB = I and BA=I. (ii) The ij th entry of the product AB is c ij =. A = [a ij] and B = [b ij] be two diagonal n? (i) Begin your proof by letting. Remark When … Proof: First observe that the ij entry of AB can be writ-ten as (AB) ij = Xn k=1 a ikb kj: Furthermore, if we transpose a matrix we switch the rows and the columns. Remember AB=BA, which means AB - BA = 0. Issues: 1. Remark Not all square matrices are invertible. If A is an elementary matrix and B is an arbitrary matrix of the same size then det(AB)=det(A)det(B). For the product AB, i) I already started by specifying that A = [aij] and B = [bij] are two n x n matrices ii) and I wrote that the ijth entry of the product AB is cij = ∑(from k=1 to n of) aik bkj Now the third part (and the part I'm having trouble with) says to evaluate cij for the two cases i ≠ j and i = j. Proof 4: Assumptions: AB = BA Need to show: A and B are both square. So det(A)det(B) = det(B)det(A) regardless of whether or not AB=BA.So if A and B are square matrices, the result follows from the fact det (AB) = det (A) det(B). I hope that helps. Theorem: Let A and B be matrices. Then if A is non singlar and I replace B with A^-1 and since we know that AB = I, then A is invertible. So det(A) and det(B) are real numbers and multiplication of real numbers is commutative regardless of how they're derived. You should be able to finish the proof,no problem now. Proof. Same goes if you if reversed then you will arrive that A and B are both invertible. Since AB is de ned, n = p. Since BA is de ned, q = n. Therefore, we have that B is n m. Thus, AB is m m BA is n n Therefore, AB and BA are both square, so we’re done. Once the linear system is in reduced row echelon form, you will see the conditions for AB=BA. Now you can set up and solve for a linear system using elementary row operations. Indeed, consider three cases: Case 1. (AB)T = B TA . This is a correct proof! as if they were x1, x2, x3, etc. If two matrices commute: AB=BA, then prove that they share at least one common eigenvector: there exists a vector which is both an eigenvector of A and B. We want to treat a,b,c, etc. Then I choose A and B to be square matrices, then A*B = AB exists. Theorem. Matrix inverses Recall... De nition A square matrix A is invertible (or nonsingular) if 9matrix B such that AB = I and BA = I. If A is invertible, then its inverse is unique. n matrices. (iii) Evaluate the entries c ij for the two cases i? 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