# eigenvalues and basis of F(R,R)

• Jan 20th 2010, 04:19 PM
234578
eigenvalues and basis of F(R,R)
So the question says to find all real eigenvalues and for each of these real eigenvalues find a real eigenfunction. V is the linear space of all functions
R->R.
a) T(f(x))=2f(x) now I can already see the only eigenvalue is 2 and any function with real coefficients works so f(x)=sinx is one such eigenfunction.

b) T(f(x))=f(-x)
Now the way I would usually approach this is I would find the matrix A=[T] with respect to the standard ordered basis of V and then use the characteristic polynomial f(t)=det(A-tI) BUT I can't figure out what the standard ordered basis would be. a basis for all polynomials would be (1, x, x^2,...) but in V, the space of ALL functions from R to R, functions like sinx are included so how can I find the matrix representation of T?

I know, for example, -1 and 1 are eigenvalues because T(sinx)=-sinx and T(cosx)=cosx but what about some function g(x)={1, when x>or=0 and g(x)={4, when x<0 ? wouldnt T(g(x))=4g(x) so can't any number be an eigenvalue of T?

Also T(h(x))=-(1/3)x if h(x)=abs(x)+2x so -1/3 is also an eigenvalue for T...so an any number be an eigenvalue? how do I show that?
• Jan 20th 2010, 05:42 PM
Jose27
Quote:

Originally Posted by 234578
So the question says to find all real eigenvalues and for each of these real eigenvalues find a real eigenfunction. V is the linear space of all functions
R->R.
a) T(f(x))=2f(x) now I can already see the only eigenvalue is 2 and any function with real coefficients works so f(x)=sinx is one such eigenfunction.

b) T(f(x))=f(-x)
Now the way I would usually approach this is I would find the matrix A=[T] with respect to the standard ordered basis of V and then use the characteristic polynomial f(t)=det(A-tI) BUT I can't figure out what the standard ordered basis would be. a basis for all polynomials would be (1, x, x^2,...) but in V, the space of ALL functions from R to R, functions like sinx are included so how can I find the matrix representation of T?

I know, for example, -1 and 1 are eigenvalues because T(sinx)=-sinx and T(cosx)=cosx but what about some function g(x)={1, when x>or=0 and g(x)={4, when x<0 ? wouldnt T(g(x))=4g(x) so can't any number be an eigenvalue of T?

Also T(h(x))=-(1/3)x if h(x)=abs(x)+2x so -1/3 is also an eigenvalue for T...so an any number be an eigenvalue? how do I show that?

For the first one you're supposed to find functions such that $2f(x)=af(x)$ where $f\neq 0$ and $a\in \mathbb{R}$ which will give you $a=2$ so you're right.

For the second one, remember that matrix representations only work for finite dim. spaces and yours isn't. Now you want $f(x)=af(-x)=a^2f(x)$ and so $a=1,-1$ so for $a=1$ you have all even functions as eigenvectors and for $a=-1$ all odd functions.
• Jan 20th 2010, 07:35 PM
234578
Quote:

Originally Posted by Jose27
For the first one you're supposed to find functions such that $2f(x)=af(x)$ where $f\neq 0$ and $a\in \mathbb{R}$ which will give you $a=2$ so you're right.

For the second one, remember that matrix representations only work for finite dim. spaces and yours isn't. Now you want $f(x)=af(-x)=a^2f(x)$ and so $a=1,-1$ so for $a=1$ you have all even functions as eigenvectors and for $a=-1$ all odd functions.

thanks for the reply, could you possibly clarify some things I didnt understand? how did you get f(x)=af(-x)=a^2f(x) ? is it because we know analytically that the function must be either even or odd?
• Jan 20th 2010, 07:55 PM
Jose27
Quote:

Originally Posted by 234578
thanks for the reply, could you possibly clarify some things I didnt understand? how did you get f(x)=af(-x)=a^2f(x) ? is it because we know analytically that the function must be either even or odd?

Apply your operator twice, or use a change of variables and apply the property that f is an eigenvector. Or its easy to see that your whole space is the direct sum of the subspace of even functions and the subspace of odd functions