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.





Now, I've already proved that the dimension of the kernel is 0 and therefore L is one-to-one. But how do I show that it is onto? If someone could explain this, it would be greatly appreciated!

Linear algebra question?
To show onto, let an element p of P_n be given. If you can explicitly find a q such that L(q) = p, then L is onto.





If I'm following the notation correctly, P_n = polynomials of highest degree n. So p is some polynomial





a_n x^n + a_(n-1) x^(n-1) + ... + a_1 x + a_0





and you want another polynomial,





b_n x^n + b_(n-1) x^(n-1) + ... + b_1 x + b_0





so that this polynomial plus its derivative equals p. Well that means that





b_n x^n + b_(n-1) x^(n-1) + ... + b_1 x + b_0 + n*b_n x^(n-1) + (n-1)*b_(n-1) x^(n-2) + ... + b_1 = a_n x^n + a_(n-1) x^(n-1) + ... + a_1 x + a_0





(we are forcing this to be true)





now just figure out what the coefficients should be. the a_i's are fixed, so you can only choose the b_i's.





b_n x^n + (b_(n-1)+n*b_n) x^(n-1) + ... + (b_1+2*b_2) x + (b_0 + b_1) = a_n x^n + a_(n-1) x^(n-1) + ... + a_1 x + a_0





so b_n = a_n,


b_(n-1) + n*b_n = a_(n-1)


.


.


.


b_1 + 2*b_2 = a_1


b_0 + b_1 = a_0








and you're going to get this system of equations, which is going to be a little messy but it should be doable. Try working it out with n=3 or n=2 if you get stuck, just to see how the computations will go at least.





Anyway hope this helps; this is just a rough outline of how the onto part would go.