Let V be the vector space of polynomial functions of degree three or less. Let the ordered basis for V consisting of the functions of the four functions xj: j = 0, 1, 2, 3 and let D be the differentiation operator. Then the matrix of D in the above ordered basis isa)b)c)d)Correct answer is option 'A'. Can you explain this answer?

Question

Answer

Let V be the vector space of polynomial functions of degree three or less. Let the ordered basis for V consisting of the functions
x^j : j = 0 ,1 ,2 , 3
and let D be the differentiation operator. We need to find the matrix of differentiation operator D in the basis x^j, j = 0,1, 2, 3, that is {1, x, x², x³}.
Therefore, D(1) = d/dx(1) = 0
= 0.1 + 0.x + 0.x² + 0.x³
D(x) = d/dx (x) = 1 = 1.1 + 0 . x + 0 .x² + 0.x³ ax
D(x²) =d/dx(x²) = 2x = 0.1 + 2.x + 0.x² + 0.x³ ax
D(x³) = d/dx (x³) = 3x² = 0.1 + 0.x + 3.x² + 0.x⁴
Thus, the matrix of differentiation operator

is

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