# Prove that v is an eigenvector of operator B

1. Jun 13, 2013

### Smazmbazm

1. The problem statement, all variables and given/known data

Let B be the linear operator

$(1-x^{2}) \frac{d^2}{dx^2}-x\frac{d}{dx}$

Show that

$T_{4}(x) = 8x^{4} - 8x^{2} + 1$

is an eigenvector of B, and find the corresponding eigenvalue.

Attempt

Righto, I find these rather difficult so a step by step solution would be nice but I know that's a lot of typing and formatting so I'm grateful for any corrections.

Step 1: $Bv_{4} = \lambda _{4} v_{4}$

Step 2: $((1-x^{2}) \frac{d^2}{dx^2}-x\frac{d}{dx})(8x^{4} - 8x^{2} + 1) = \lambda _{4} (8x^{4} - 8x^{2} + 1)$

Step 3: This is the first point where I'm unsure of what comes next. Should I be doing something like this

$\frac{d^2}{dx^2}((1-x^{2})(8x^{4} - 8x^{2} + 1)) - x\frac{d}{dx}(8x^{4} - 8x^{2} + 1) = 8\lambda _{4}x^{4} - 8\lambda _{4}x^{2} + \lambda _{4} \\ \frac{d^2}{dx^2}(-8 x^6 +16 x^4 -9 x^2 + 1) - \frac{d}{dx}(8x^{5} - 8x^{3} + x) = 8\lambda _{4}x^{4} - 8\lambda _{4}x^{2} + \lambda _{4} \\ -240x^4 + 192x^2 - 18 - 40x^4 + 18x^2 - 1 = 8\lambda _{4}x^{4} - 8\lambda _{4}x^{2} + \lambda _{4} \\ -280x^4 + 216x^2 - 19 = 8\lambda _{4}x^{4} - 8\lambda _{4}x^{2} + \lambda _{4}$

Is this all correct so far? If so, awesome. Now I don't really know how to go about solving for $\lambda$

In general, is it enough to say that $\lambda = \frac{-280x^4 + 216x^2 - 19}{8x^{4}- 8x^{2} + 1}$

Last edited: Jun 13, 2013
2. Jun 13, 2013

### CompuChip

You can do that, but you don't need the right hand side. It's probably easier to start by calculating
$$B v = \frac{d^2}{dx^2}((1-x^{2})(8x^{4} - 8x^{2} + 1)) - x\frac{d}{dx}(8x^{4} - 8x^{2} + 1)$$
and after simplifying you should get a multiple of $8x^{4} - 8x^{2} + 1$ from which you can read off the value of $\lambda$.

No! You cannot move the x after the derivative operator:
$$x \frac{d}{dx} x^n \neq \frac{d}{dx} x^{n+1}$$

3. Jun 13, 2013

### Smazmbazm

Ok great, thanks for the response. By that logic, wouldn't it imply that you can't move the $(1-x^2)$ to the right side either? So you just take the double derivative of the eigenvector then multiply by $(1-x^2)$ minus the first derivative of the eigenvector multiplied by x then solve for lambda. If I follow that process, I get a lambda of -16 =] so I guess that's the way to do it