5.2 The Characteristic Equation Calculation of determinants Suppose A is an nn× matrix and that matrix U is any echelon form obtained from A by row replacements and row interchanges with no scaling. Let r be the number of such row interchanges. r Then the determinant of A is (−1) times the product of the main r diagonal elements in U. Thus Det( A ) = (−1) u11 u 22 unn. r Det() A= (−1) u11 u 22 unn. If A is invertible, then uu11,, 22 , unn are all pivots (since AIn and the uii have not been scaled so that they are all 1's) Otherwise, at least unn is zero, and then the product uu11 22 unn =0. To summarize: r n (−1) ∏uAii is invertible Det() A = i=1 0 A is not invertible Theorem Let A be an nn× matrix. Then A has an inverse if and only if: 1. All eigenvalues of A are nonzero. 2. The determinant of A is not zero. Theorem (Properties of determinants) Let A and B be nn× matrices. 1. A is invertible ⇔≠Det( A) 0. 2. Det( AB) = ( Det( A))⋅( Det( B)). 3. Det( AT ) = Det( A). 4. If A is triangular, then Det( A) is the product of the main diagonal entries of A. 5. A row replacement operation on A does not change the determinant. A row interchange changes the sign of the determinant. A row scaling also scales the determinant by the same scalar factor. r Det() A= (−1) u11 u 22 unn. If A is invertible, then uu11,, 22 , unn are all pivots (since AIn and the uii have not been scaled so that they are all 1's) Otherwise, at least unn is zero, and then the product uu11 22 unn =0. To summarize: r n (−1) ∏uAii is invertible Det() A = i=1 0 A is not invertible Matrix similarity Let A and B be n× n matrices. A is said to be similar to B if there exists an invertible matrix P such that −− P11 AP= B or equivalently, A= PBP . −− If we write Q for P11, we have Q BQ= A. So B is similar to A as well. Thus AB and are simply said to be similar. Theorem If nn× matrices A and B are similar, then they have the same characteristic polynomial and thus the same eigenvalues. End presentation .