Linear algebra question?

Now I just attempted the following proof:





Show that L: P_n -%26gt; P_n given by L(p) = p+p' is an isomorphism. First show that L is a linear operator. He said to do it this way:





Suppose p and q are in P_n. Then:


L(p+q) = p+q+(p+q)'


= p+p'+q+q'


= L(p)+L(q)


L(c*p)=c*p+(c*p)'


=c*p+c*p'


=c*(p+p')


=c*L(p)


So L is a linear operator.





Suppose that p is in the kernel of L. Then L(p) = 0. By definition of L, this means that p'=-p. The only element of P_n that would satisfy this equation is 0. This is because the derivative operator always lowers the degree of a polynomial by 1. Since the dimension of the kernel is 0 and P_n is finite dimensional, L is an isomorphism.





Now, I was wondering if someone could show me how I would prove that L is onto (because this appears to be missing in my proof).

Linear algebra question?
Let q be an arbitrary polynomial. q=(a_n)(x^n)+(a_n-1)(x^n-1)+ . . . .+(a_0). Now let p=(b_n)(x^n)+(b_n-1)(x^n-1)+ . . . .+(b_0). Then L(p)=p+p'=(b_n)(x^n) + ((b_n-1) +(n*b_n)(x^n-1) . . . . .(a_0 +a_1) if this equals q we need b_n = a_n so we defined what b_n is.





next look at the x^n-1 term . if L(p) still equal q we need (b_n-1) +(n*b_n)=a_n-1 thus we define b_n-1 this way. etc we define b_n-2, b_n-3 etc. so we found a polynomial p. s.t. L(p)=q