Question

1. Let T: Pn(R) + Pn+1(R) be defined: T(P(x)) = (x + 1)p(x + 2) (c) (8 marks) Find a matrix representation for T with respect to the standard bases {1, 2, ...,2"} for Pn and {1, 2, ...,xN+1} for Pn+1 if n = 4. (d) (5 marks) Let D : Pn+1(R) + Pn(R) be the derivative operator. What is the rank of DoT? Justify your answer. Describe ker(DoT). Is DoT one-to-one? (e) (5 marks) What is the rank of ToD? Justify your answer. Describe ker(ToD). Is To D one-to-one?

Answer 1

1c) het Ti P(K) a Par (K) be defined T($90) = (x+12 (x+2) when n=4 for Theo T(1) = (2+1) - 1x T(X)= (x+12 (2+2) = 2+ 3x + x2 162) = (x+1) (2+2)? = 4+ out 5x² + x3 T(2²) = (x+1) (x+233 = 8+202 + 18x2 +373 txt 1(24) = (x+1)(x+2)4 = 16+48x + 56x2 +32x3 +9x4+x5

The matrix representation for I want hy to P since n=4. 2 4 8 16 - 3 B 20 48 A- - 0 5 18 36 0 0 ។ 32 - 0 0 0 9 6 D - 0 d) het D Pat (R) 4 (R) is the derivative operator: Dlbląg - Se (plus) D(= DW = 1 D (x²) = 2x LD (x²) = 3x² (x4) = 4x² Lox) = 5x4 The matrix representation for in Lo o 0 0 0 0 O 0 6 2 3 0 0 Bl= 0 0 0 4 0 0 0 0 pin 0 0 0 6 0 BA Now of the marrix yor DoT in

& 20 48 BA= 0 jo 36 O 0 96 0 0 0 36 0 O 0 O since it u the upper triangulae mateix mith positive non-zuia entries. BA U a full rank mattix. rank of مك Do I 5.2 non-zus ro M KeL(DoT) since BA (matrix of DoT) is full rank mateix Thue i no pivot column so Kee (Do I) has only trivial solution. 0 0 Kee (DoT) = DoT i one-to-one Since each column containe a pivot element, So, DOT i une - J - me.

The matrix for ToD in AB 1 4 32 o o 80 75 - 6 24 20 240 M AB = 0 0 2 15 12 280 € o 3 Oo 20 ooo 4 Oo 45 5 6 0 이 + The reduced 21 72 echelon form of AB is 13 24 us 0 0 o Non 0 0 O 0 0 0 0 o 0 Zero teftab 0 0 0 O Rows GOO 0 Co O o Oo oo 0 0 Rank ToD = og rank of AB = 1 number of non-zu would in theftAB) Rank of TOD 5 Ker(TOD) Null space of AB (AB)x=0 = = xyz xs-x6=0 we get then

H11 x 0 13 0 0 " xy 0 o o κς 126 0 Ker (ToD)=9 0 1 Now since column I pivot element one: does not contain To D u not one- So,