# Vector space

1. Nov 10, 2007

### smoothman

any ideas on how to go about conducting these please. i will attempt them once i have a clear idea on how to do this. thanx :)

let V be the vector space of polynomials over C of degree <= 10 and let
"D: V -----> V" be the linear map defined by

D(f) = df/dx

show
(1) D^11=0
(2) deduce 0 is the only eigenvalue of D
(3) find a basis for the generalised eigenspaces v1(0), v2(0), and v3(0).

2. Nov 10, 2007

### d_leet

What have you tried already? (1) is very easy, just pick an arbitrary degree 10 polynomial with coefficients in C and take 11 derivatives.

3. Nov 11, 2007

### matt grime

(1) is very easy, *don't* pick an arbitrary degree ten poly and take 11 derivatives. That would be tedious and missing the point: what happens to the degree of a polynomial if you differentiate it?

4. Nov 11, 2007

### smoothman

well ur right there matt..
if u differentiate a polynomial then its degree reduces by 1...
so if u differentiate 11 times, a polynomial of degree 10, then the answer would be 0 as stated in the question.. but the problem is .. HOW can i show this mathematically?
thnx.

5. Nov 11, 2007

### CompuChip

I'd just do it like this; Suppose P is a polynomial of degree d < 11. Then
$$P = \sum_{n = 0}^d a_n x^n$$
where any coefficient may be zero (in particular, if $a_d = a_{d - 1} = \cdots = a_{k + 1} = 0, a_k \neq 0$ we have a polynomial of degree k.
Differentiation is linear, so
$$DP = \sum_{n = 0}^d a_n D x^n = \sum_{n = 1}^d n a_n x^{n-1} = \sum_{n = 0}^{d - 1} b_n x^n$$
where $b_n = n a_n$, which is the general form for a polynomial of degree d - 1 or lower.

Or what about induction :)
The degree is the highest power that occurs.
A polynomial of degree 0 is a constant, differentiation makes it zero.
A polynomial of degree 1 is of the form $$c x + P_0$$ with $P_0$ of degree zero, differentiating gives $$c[/itex] which is of degree zero, so the degree is lowered by one. Now suppose it is true for degrees up to and including n. Suppose $P_{n+1}$ is a polynomial of degree n + 1. Then the coefficient of the x^(n+1) term is non-zero by definition, so we can divide it out and write $P_{n+1} = x^{n+1} + P_n$, with P_n a polynomial of degree n. Since D is linear, $D P_{n+1} = D x^{n + 1} + D P_n = (n + 1) x^n + D P_n$, where the first term is a polynomial of degree n and the second term is of degree strictly smaller than n by the induction hypothesis. The sum of two polynomials has no greater degree than the degree of the largest, so the degree is maximally n < n + 1. But maybe matt has a faster way in mind. 6. Nov 11, 2007 ### smoothman thnx compuchip.. that method looks way too complicated.. especially for the first part of a question. any quicker methods please? also.. how would you go about conducting part (2) and (3) of the question please? thnx a lot :) 7. Nov 11, 2007 ### HallsofIvy Matt Grime gave a simple explanation and you asked for a more "mathematical" argument. Comuchip gave you two rigorous calculations and you complain that they are "way too complicated"! This YOUR problem. Find an argument you are happy with. In (2), suppose there were some eigenvalue of D, $\lambda\ne 0$. Let f be an "eigenfunction" corresponding to $\lambda$. Then $Df= \lambda f$ and $D^2 f= D(\lambda f)= \lambda^2 f$. What is D11 f? As for (3) if you look up the definition of "generalized eigenspace" I think you'll find that a very easy question! 8. Nov 11, 2007 ### CompuChip smoothman, would you like an argument like the following better? [tex] \begin{array}{l} D^0 f=a_{10} x^{10}+a_9 x^9+a_8 x^8+a_7 x^7+a_6 x^6+a_5 x^5+a_4 x^4+a_3 x^3+a_2 x^2+a_1 x+a_0 \\ D f=10 a_{10} x^9+9 a_9 x^8+8 a_8 x^7+7 a_7 x^6+6 a_6 x^5+5 a_5 x^4+4 a_4 x^3+3 a_3 x^2+2 a_2 x+a_1 \\ D^2 f=90 a_{10} x^8+72 a_9 x^7+56 a_8 x^6+42 a_7 x^5+30 a_6 x^4+20 a_5 x^3+12 a_4 x^2+6 a_3 x+2 a_2 \\ D^3 f=720 a_{10} x^7+504 a_9 x^6+336 a_8 x^5+210 a_7 x^4+120 a_6 x^3+60 a_5 x^2+24 a_4 x+6 a_3 \\ D^4 f=5040 a_{10} x^6+3024 a_9 x^5+1680 a_8 x^4+840 a_7 x^3+360 a_6 x^2+120 a_5 x+24 a_4 \\ D^5 f=30240 a_{10} x^5+15120 a_9 x^4+6720 a_8 x^3+2520 a_7 x^2+720 a_6 x+120 a_5 \\ D^6 f=151200 a_{10} x^4+60480 a_9 x^3+20160 a_8 x^2+5040 a_7 x+720 a_6 \\ D^7 f=604800 a_{10} x^3+181440 a_9 x^2+40320 a_8 x+5040 a_7 \\ D^8 f=1814400 a_{10} x^2+362880 a_9 x+40320 a_8 \\ D^9 f=362880 a_9+3628800 x a_{10} \\ D^{10} f=3628800 a_{10} \\ D^{11} f=0 \end{array}$$

Last edited: Nov 12, 2007
9. Nov 11, 2007

### matt grime

Eugh. Differentiation is linear, so it suffices to consider only monomials.

10. Nov 12, 2007

### theperthvan

that's really quite elephant, I mean elegant