Show that the only eigenvalue of the map D(f) = (d/dx)f is 0. Is D diagonalisable? and why?
I'm gonna assume this is a linear operator on a finite dimensional polynomial space.
so basically what this question is asking is, If you have a polynomial P(x) = $\displaystyle a_nx^n+...+a_0$ then $\displaystyle d\dx(P(x)) = n*a_nx^{n-1} +... + a_1$
when is applying this operator on a polynomial equal to multiplying the polynomial by a constant (eigen value)? Notice that d/dx reduces the degree of the polynomial if you have a non constant polynomial, so can you ever do $\displaystyle \lambda * P(x) = d/dx(P(x))$? If P(x) is non constant (meaning it has atleast 1 x term) then the left side would be of a higher degree than the right side. So this must possible only when p(x) is constant. when p(x) is constant then d/dx p(x) = 0. so $\displaystyle p(x) = a_0 $ $\displaystyle d/dx(p(x)) = 0 = 0*p(x)$ so 0 is the only eigen value of the map. The eigen vectors are all the constant polynomials in the polynomial space.
A linear map is diagnizable iff the sum of dimensions of its eigen spaces is equal to dim V. So basically can we find n lin. independent eigen vectors for the linear map? If so then the linear map is diagnizable. Since we only found once eigen vector for the linear map, this matrix is not diagnizable. (Assume the polynomial space is not just a bunch of constants i.e has dim > 1).
thanks for your reply
basically what you wrote is what is what is intuitive, i can observe that result quite simply. but i have to 'show' it, i think that means some i require some technical reasoning. i agree with what you wrote quite simply.