Question

is an eigenvalue invertible matrix with X as an eigenvalue. Show that of A-1. Suppose v ER is a nonzero column vector. Let A (a) Show that v is an eigenvector of A correspond zero column vector. Let A be the n xn matrix vvT. n eigenvector of A corresponding to eigenvalue = |v||2. lat O is an eigenvalue of multiplicity n - 1. (Hint: What is rank A?) (b) Show that 0 is an eigenvalue of

Answer 1

Air ioverhble mcahrix AS 4 i eigenvcalue There cxirt e eigen rector sulh thau Ax=1X Airinverhtle mulhpy both side 61 A AAXACAz) As A AX AAK A'M=GJ J in elgen value ct A VEIRA 10 a non zem caloump vecho V1-- A V. VT Connder Avvv TvE - =V.(VT. v) = viv2 \\vu2 ) Av INV Vis eigenrecho at A comerpendiy to eige ralue A=v VT vivn AS V2 Vn V VI Vo Vo VOVi- Jt we tauee tauo column at A then they are linear J dependeni Tank (A) i ofo But vto TOnk CA) - 1' Pant nullih th m ant At nulli A numberT a coloumn

+nuti'y ulliy A= n-1 A H set se eipAxo has n- dimen hon o js eigenvedue cut A with mulhpliuy liv is eigenval ue ot A coith mulip luy f