Eigenvalues of linear operators

[LocationX]

Let V be the vector space of all real integrable functions on [0,1] with inner product <f,g>=\int_0^1 f(t)g(t)dt

Three linear operators defined on this space are A=d/dt and B=t and C=1 so that Af=df/dt and Bf=tf and Cf=f

I need to find the eigenvalues of these operators:

For A:
\frac{df}{dt} = \lambdaf
\frac{d ln(f)}{dt} = \lambda
d(ln(f))=\lambda dt+c
f=ce^{\lambda t}

So the eigenvalues for A are continuous and can be any real number.

For B:
tf=\lambda f
\lambda =t

The eigenvalues are the variable t?

For B;
Cf=f=\lambda f
The eigenvalue is 1.

Confused about B.

 

[HallsofIvy]

Let V be the vector space of all real integrable functions on [0,1] with inner product <f,g>=\int_0^1 f(t)g(t)dt

Three linear operators defined on this space are A=d/dt and B=t and C=1 so that Af=df/dt and Bf=tf and Cf=f

I need to find the eigenvalues of these operators:

For A:
\frac{df}{dt} = \lambdaf
\frac{d ln(f)}{dt} = \lambda
d(ln(f))=\lambda dt+c
f=ce^{\lambda t}

So the eigenvalues for A are continuous and can be any real number.

I'm not sure I like the statement "the eigenvalues of A are continuous". What does it mean for a number to be "continuous"? But, yes, every real number is an eigenvalue- the set of eigenvalues is a continuous interval rather than discreet.

For B:
tf=\lambda f
\lambda =t

The eigenvalues are the variable t?

No, eigenvalues are NUMBERS, not variables. tf= \lambda f if and only if f(t)= 0 for all t, the "zero vector". This operator has NO eigenvalues.

For B;
Cf=f=\lambda f
The eigenvalue is 1.

Confused about B.