MAT224 ASSIGNMENT 2 SOLUTIONS DUE BY FRIDAY JUNE 5, 2020, 11:59 PM
Each question is worth 5 marks.
Question 1. Define T : Pn → R via T(p(x)) = ak, where p(x) = a0+a1x+…+anxn. Prove that dim(kerT) = n.
Solution. Notice that image of T is R (you need to prove this). Hence, rank(T) = 1. By the Dimension Theorem, n+1=dimPn =dim(kerT)+rank(T)=dim(kerT)+1.
Question2.LetV→T W→S VbefunctionssuchthatTS=1W andST=1V.IfTislinear,provethatSislinear. Solution. Given w1, w2 ∈ W and a ∈ R,
and
is linear, and that ker T = {0} if and only if {v1, . . . , vn} is an independent set. Solution. Given (a1,…,an),(b1,…,bn) ∈ Rn, and c ∈ R,
and
T((a1,…,an)+(b1,…,bn))
= T(a1 +b1,…,an +bn) n
= (ak+bk)vi k=1
nn
= akvk+bkvk k=1 k=1
= T(a1,…,an)+T(b1,…,bn),
n k=0
S(w1 + w2) =
= S(T (S(w1) + S(w2))) since T is linear
S(TS(w1) + TS(w2)) since TS = 1W = S(w1)+S(w2) since ST = 1V ,
S(aw1) = S(a(T S(w1))) = S(T (aS(w1))) = aS(w1).
n
Question 3. Given {v1,…,vn} in a vector space V, define T : Rn → V by T(a1,…,an) = akvk. Show that T
n
T(c(a1,…,an)) = T(ca1,…,can) = (cak)vk = cakvk = cT(a1,…,an).
k=1 k=1
n
Hence, T is linear. Notice that ker T = {0} if and only if T is one-to-one. Suppose T is one-to-one and civi = 0.
k=1 Then T(c1,…,cn) = 0 = T(0,…,0) ⇒ ci = 0 for i = 1,…,n. Hence, {v1,…,vn} is an independent set.
Conversely, suppose {v1, . . . , vn} is an independent set. If T (c1, . . . , cn) = 0, then
n
civi = 0 ⇒ ci = 0 for all i = 1,…,n ⇒ (c1,…,cn) = (0,…,0) ⇒ kerT = {0}. k=1
Question 4. Suppose T : P2 → P3 is a linear transformation with T(x2) = x3, T(x + 1) = 2, and T(x − 1) = 2x. Find the following.
(a) T (x2 + 2x + 2).
Solution. T(x2 +2x+2)=T(x2)+2T(x+1)=x3 +4.
1
k=1
2 MAT224 ASSIGNMENT 2 SOLUTIONS DUE BY FRIDAY JUNE 5, 2020, 11:59 PM
(b) T(x).
Solution.T(x)=T 2(x+1)+2(x−1) =2(T(x+1)+T(x−1))=2(2+2x)=1+x.
Question 5. Suppose N is an n × n matrix and Nk = 0 (the zero matrix) for some natural number k. Prove that T :Mnm →Mnm givenbyT(A)=A−NAisanisomorphism.
Solution. For any X,Y ∈ Mnm and a ∈ R,
T(X+Y) = (X+Y)−N(X+Y)=X−NX+Y −NY =T(X)+T(Y) T(aX) = aX−N(aX)=a(X−NX)=aT(X).
Hence,T isalineartransformation. Moreover,ifX∈kerT,then0=T(X)=X−NX⇒X=NX. Consequently, T(NX) = T(X) = 0 ⇒ 0 = NX − N(NX) ⇒ NX = N2X.
That is, X = NX = N2X. Continuing this process shows that X = NX = N2X = … = NkX = 0. Therefore, ker T = {0}. That is, T is one-to-one and by Theorem 7.3.3, T is an isomorphism.
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