Let Pn() be the vector space of all polynomials of degree atmost n.Let g(x) = x + 1 and define T : P2()→P2() byT(f (x)) = f(x) g(x) + 2f (x).Then the trace of A is;a)5b)6c)9d)12Correct answer is option 'C'. Can you explain this answer?

Question

Let Pn() be the vector space of all polynomials of degree atmost n.Let g(x) = x + 1 and define T : P2()&rarr;P2() byT(f (x)) = f(x) g(x) + 2f (x).Then the trace of A is;a)5b)6c)9d)12Correct answer is option 'C'. Can you explain this answer?

Answer

We know that {1, x, x²} is the standard basis for P₂ (ℝ)

Now, T (1) = 0 - g(x) + 2·1 = 2

T(x) = 1(x + 1) + 2x =3x + 1

T(x²) = 2x(x + 1) + 2x² = 2x² + 2x + 2x² = 4x² + 2x

⇒ The matrix of T with respect to standard basis is A =

Clearly.A is a triangular matrix

So eigen values of A are diagonal entries of A

⇒ Eigen values of A are 2,3,4 ⇒ Tr (T) = Tr(A) = 2 + 3 + 4 = 9

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