![]() I would guess the assertion is also true if $K$ is finite, but I don't know how to show it. ![]() Moreover, determinants are used to give a formula for A1 which, in turn, yields a formula (called Cramer’s rule) for the solution of any. One consequence of these theorems is that a square matrix A is invertible if and only if det A 60. ![]() For any two nxn invertible matrices U and V, the inverse of UV is (UV)-10. 3.2 Determinants and Matrix Inverses In this section, several theorems about determinants are derived. UV is invertible because the product of any two matrices is always invertible. The product of two invertible matrices is also invertible. So I'll prove the claim for $|K|=\infty$. Since U and V are orthogonal, each is invertible by the definition of orthogonal matrices. ![]() Remark When A is invertible, we denote its inverse as A' 1. If A is invertible, then its inverse is unique. (We say B is an inverse of A.) Remark Not all square matrices are invertible. Initially my assumption was that the question was primarily about finite fields $K$, but after reading the comments it appears to me that there is some confusion about the infinite case as well. DeÞnition A square matrix A is invertible (or nonsingular ) if matrix B such that AB I and BA I.
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |