Book
Linear Algebra
Author
Gilbert Strang
Edition
A=A0=matrix([[4,0,0],[0,5,0]]) A r=m=A.rank() r C=(A.transpose()*((A*A.transpose()).inverse())) C var('b13,b23') B=matrix([[1/4,0,b13],[0,1/5,b23]]) B B*A.transpose() T1=matrix([[4,0],[0,5],[0,0]]) T1 A Arank1=matrix([[2,1,1],[4,2,2],[8,4,4],[-2,-1,-1]]) Arank1.rank() u=matrix([[1],[2],[4],[-1]]) vt=matrix([2,1,1]) Res=u*vt Res Res==Arank1
Solution by:
<T.R.N.Karthik>, <Student>, <Gitam University>