Let P3 denote the vector space of all polynomials of degree at most 3. Briefly explain why the derivative is a linear transformation from P3 to P3. We will denote this transformation by D. Show that the only eigenvalue of D : P3 P3 is zero, and describe the eigenspace for this eigenvalue. Solution a): transformation is denoted by D; is defined as follow ........ D(ax^3 +bx^2 +cx +d) = 3ax^2 +2bx +c. now take ........,ANY TWO v1 , v2....belongs to P_3 . v1 = a1x^3 +b1x^2 +c1x +d1. v2 = a2x^3 +b2x^2 +c2x +d2. k*v1+v2 = (k*a1+a2)x^3+(k8b1+b2)x^2+(k*c1+c2)x+(k*d1+d2). now; D(k*v1+v2) = 3 (k*a1+a2)x^2 + 2(k*b1+b2)x+(k*c1+c2) = ( 3(k*a1)x^2 +2(k*b1)x+k*c1 ) + ( 3(k*a2)x^2 +2(k*b2)x+k*c2 ) = k*D(v1)+ D(v2). this proves that D is a linear transformation. b): let ; v be eigenvector with an eigenvalue = t. then; D(v) = tv D(ax^3 +bx^2 +cx +d) = t(ax^3 +bx^2 +cx +d) 0*x^3 + 3ax^2 +2bx +c = t*ax^3 +t*bx^2 +t*cx +t*d implies that ...... t*a = 0 t = 0. so; only eigen value is zero for constant polynomial. good luck!!!!!!!!!! .