tag:blogger.com,1999:blog-2237906900456925233.post6891473311312437483..comments2023-10-18T23:36:58.396+08:00Comments on 數學資料庫手記: Some linear algebraMathDBhttp://www.blogger.com/profile/18353093974880979999noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-2237906900456925233.post-83202372323842846682009-05-23T04:15:15.854+08:002009-05-23T04:15:15.854+08:00Nice post Polam! In (1), you prove that AB and BA ...Nice post Polam! In (1), you prove that AB and BA have the same characteristic polynomials by approximating A with a sequence of invertible matrices. I think there is another way of proving this without using limit. Simply observe that the characteristic polynomial of A is <br />\det(A-tI)=\sum_{i=1}^n(-1)^i\text{tr}(\bigwedge^{n-i}A)t^i<br />where \bigwedge^i A:\bigwedge^i C\to\bigwedge^i C is given by <br />\bigwedge^i A(z_1\wedge\cdots\wedge z_i)=Az_1\wedge\cdots\wedge Az_i<br /><br />and \bigwedge^i(AB)=\bigwedge^i A\bigwedge^i B. So the equality of the characteristic polynomials of AB and BA boils down to the fact that tr(CD)=tr(DC).Alexhttps://www.blogger.com/profile/12583638980972443433noreply@blogger.com