Question

Q2 15 Points Let A € Mnxn(R). Define trace(A) = {1-1 Qiji (i. e. the sum of the diagonal entries) and tr : Mnxn(R) +R, A H trace(A). Q2.1 2 Points Show that U = {A € Mnxn(R): trace(A) = 0} is a subspace of Mnxn (R). Please select file(s) Select file(s) Q2.2 4 Points Compute dim(im(tr)) Enter your answer here and dim(ker(tt) Enter your answer here each (1pt) Justify your answer. (2pt) Enter your answer here Q2.3 5 Points Find a basis for ker(tr) and verify that it is in fact a basis. Please select file(s) Select file(s) Q2.4 3 Points Show that for any A € Mnxn(R) there is B e ker(tr) and a € R such that A = B +a. E for E E Mnxn(R) with (E)i,j Si ifi =j=1 otherwise i. e. E is the matrix with 1 in the (1, 1)-entry and everywhere else. Please select file(s) Select file(s) Q2.5 1 Point What is a E R from part (4) equal to? Enter your answer here

Answer 1

trace () = Now, trace (A+B) = trace (A) + trace (B) = 0 to=0 Solution: Q: 2 Given, AE Mnxn n (IR) aji and define to: Mnxn (IR). → IR by A trace (A) Q: 2.1 Given, u = {AE Mnxn (IR) : trace (A) = claim uis a subspace of O EU, because, trace (Zero matrix) = 0 » ut o Let, A, B eu [ Assume A,BE Mnxn (IR)] trace (A)=0 and trace (B)=0 o Maxn (IR). - A+BE U. (i) Let LEIR, and, A EU trace (2A) Now, a trace (A) = 2.0=0 H 2 A E u From () , (ii) and (iii) We conclude That wisa subspace of im (to(a)) = { trace (A): A E Mnxn (IR)} im (too ) (A) = trace (A) Mnxn (IR) Maxn (IR). 2.2 2 AE a11 + 922+ tann since, all + a22t. tann ER Hence, dim (im (1)

dim (im (to)) = 1 because 1 is a basis of im (to) Now ker (to) = {A At Maxn (IR): to (A)=0} = {AE Maxi (IR) : trace (A)=0} = {n A E Maxn (IR): Eaii = 1 = 0] n i=1 since, dim (Maxn (IR)) = n² dim (Ker (to)) = n²-1 Hence,