MAT224 ASSIGNMENT 4 SOLUTIONS DUE BY FRIDAY AUGUST 7, 2020, 11:59 PM
Remark: 5 of the following questions will be marked, but you must do all of them. Each question that is marked is out of 5 for a total of 25 marks.
Question 1. Let D : P3 → P2 be the differentiation map given by D(p(x)) = p′(x). Find the matrix of D corresponding to the basis B = {1,x,x2,x3} and E = {1,x,x2} (that is, find MEB(D)), and use it to compute D(a+bx+cx2 +dx3).
Solution.
Moreover,
0003 CE(D(a + bx + cx2 + dx3))
So
MEB(D) = CE(D(1)) CE(D(x)) CE(D(x2)) CE(D(x3)) = CE(0) CE(1) CE(2x) CE(3×2)
0 1 0 0 = 0 0 2 0.
PD←B = MDB(1P3 ) = CD(1)
CD(x) CD(x2) 1
1 CD(x3) = 0
CD(a+bx+cx2 +dx3)=CD(p(x))=PD←BCB(p(x))=0 0
1 1 −1 −2 0 1
= MEB(D)CB(a + bx + cx2 + dx3) 0 1 0 0a
= 0 0 2 0 b c
0003 d b
= 2c. 3d
Hence, D(a + bx + cx2 + dx3) = b + 2cx + 3dx2 (which is what you would expect).
Question 2. In P3 find PD←B if B = {1,x,x2,x3} and D = {1,(1 − x),(1 − x)2,(1 − x)3}. Then express p(x) = a+bx+cx2 +dx3 as a polynomial in powers of (1−x).
Solution.
0 0 0 −1 d −d Hence,a+bx+cx2+dx3 =p(x)=(a+b+c+d)+(−b−2c−3d)(1−x)+(c+3d)(1−x)2−d(1−x)3.
Question 3. Define T : R2 → R2 via T (a, b) = (a − b, 2b − a). Find det T , tr T , and CT (x). Solution. Let B = {(1, 0), (0, 1)}. Then,
1 −1 MB(T) = [CB(T(1,0)) CB(T(0,1))] = [CB(1,−1) CB(−1,2)] = −1 2 .
x−1 1 2 Hence,detT =detMB(T)=1, trT =trMB(T)=3, CT(x)=det 1 x−2 =x −3x+1.
Question 4. Let T : V → V be a linear operator satisfying T2 = T. Define U1 = {v ∈ V : T(v) = v} and U2 = {v ∈ V : T(v) = 0}. Prove that V = U1 ⊕ U2.
Solution. Suppose x ∈ U1 ∩ U2. Then 0 = T(x) = x. Hence, U1 ∩ U2 = {0}. Notice that
1
1 1 1 −1 −2 −3 .
0 1 3 0 0 −1
1 a a+b+c+d −3b=−b−2c−3d. 3 c c + 3 d
0 0
2 MAT224 ASSIGNMENT 4 SOLUTIONS DUE BY FRIDAY AUGUST 7, 2020, 11:59 PM
• T(T(v))=T2(v)=T(v)foranyv∈V ⇒T(v)∈U1 foranyv∈V.
• T(v−T(v))=T(v)−T2(v)=T(v)−T(v)=0foranyv∈V ⇒v−T(v)∈U2 foranyv∈V.
In particular, for any v ∈ V ,
That is, V = U1 + U2. Hence, V = U1 ⊕ U2.
Question 5. Suppose U and W are subspaces of V , dim V = n, dim U + dim W = n, and U ∩ W = {0}. Prove that V=U⊕W.
Solution. Since U + W is a subspace of V and
dim V = n = dim U + dim W = dim U + dim W − dim(U ∩ W ) = dim(U + W ), we have U + W = V . Hence, V = U ⊕ W .
v=T(v)+(v−T(v))∈U1 +U2.