If and is supertriangular then
theorem: Let be commutative ring with identity.
If an n-square matrix
is supertriangular then .
proof: Find the characteristic polynomial of by computing the determinant of . The square matrix is a triangular matrix. The determinant of a triangular matrix is the product of the diagonal element of the matrix. Therefore the characteristic polynomial is and by the Cayley-Hamilton theorem the matrix satisfies the polynomial. That is .
|Title||If and is supertriangular then|
|Date of creation||2013-03-22 13:44:39|
|Last modified on||2013-03-22 13:44:39|
|Last modified by||Daume (40)|