# Thread: Differentiation of infinitely differentiable functions

1. ## Differentiation of infinitely differentiable functions

Let D be the differentiation on the space W of of all infinitely differentiable functions.

a. What are all the eigenvalues and eigenvectors of D?
b. Can you find a finite dimensional subspace of W of dimension 3 on which D is diagonalizable? If so, give an example of such a subspace, if not prove that it is impossible.

Where do I even start here?

2. Originally Posted by davesface
Let D be the differentiation on the space W of of all infinitely differentiable functions.

a. What are all the eigenvalues and eigenvectors of D?
b. Can you find a finite dimensional subspace of W of dimension 3 on which D is diagonalizable? If so, give an example of such a subspace, if not prove that it is impossible.

Where do I even start here?
For a., I think you're overthinking this, you're looking for $f\in C^{\infty}[\mathbb{R},\mathbb{R}]$ such that $Df=\lambda f$....or put in maybe more familiar terms $f'(x)=\lambda f(x)$. With that in mind I think it's clear.

b. What are your first intuitions?

3. Originally Posted by Drexel28
For a., I think you're overthinking this
I think you're right.

The solution to that ODE is $f(x)=e^{\lambda x}+c, c \in \mathbb{R}$, which is the general form of the eigenvectors of D. From this, and knowing that $\lambda$ must be a scalar in the space W, I would make the claim that the real numbers constitute all of the eigenvalues.

If I'm correct in both of the above claims, then it would seem as though it would be impossible to find a finite dimensional subspace of W on which D is diagonalizable because the subspace couldn't be both finite dimensional and closed under addition. How to show this though...I have no idea.

4. Originally Posted by davesface
I think you're right.

The solution to that ODE is $f(x)=e^{\lambda x}+c, c \in \mathbb{R}$, which is the general form of the eigenvectors of D. From this, and knowing that $\lambda$ must be a scalar in the space W, I would make the claim that the real numbers constitute all of the eigenvalues.

If I'm correct in both of the above claims, then it would seem as though it would be impossible to find a finite dimensional subspace of W on which D is diagonalizable because the subspace couldn't be both finite dimensional and closed under addition. How to show this though...I have no idea.
Maybe I'm misunderstanding but why doesn't $W=\text{span }\{\text{exp}(x),\text{exp}(2x),\text{exp}(3x)\}$. Then, \$\dim W=3[/tex] and all there of those are eigenvectors of $D$

5. I suppose you could pick the span of those 3 eigenvectors to be a subspace $W^'$, but that seems to present the problem of not completing W as per the definition.

Specifically, since you've taken W to be only over $\mathbb{R}$, it seems to me that sin and cos would be excluded from $W^'$ since they are complex expressions relative to that subspace.

Can you perhaps explain why taking W only over the reals doesn't lead to a loss of generality? It just seems to me that adding this restriction makes it impossible to complete W with only real numbers and the exponential function.

6. What do you mean by "completing W as per the definition"? What definition? Assuming this is a vector space over the real numbers, then taking $e^{\lambda_1 x}$, $e^{\lambda_2 x}$, and $e^{\lambda_3 x}$ for any three real numbers gives a three dimensional subspace, the subspace of infinitely differentiable functions spanned by those three functions, on which "D is diagonalizable". Specifically, on that subspace, D is represented by the matrix $\begin{bmatrix}\lambda_1 & 0 & 0 \\ 0 & \lambda_2 & 0 \\ 0 & 0 & \lambda_3\end{bmatrix}$. Unless there is more that you haven't told us, that's all the problem requires.

If this is a vector space over that complex numbers, take $e^{\lambda_1 x}$, $e^{\lambda_2 x}$, and $e^{\lambda_3 x}$ to be any three complex numbers.

7. Okay, I was misunderstanding. Thanks for the help.