Eigenvalues of linear operators

[LocationX]

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

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

I need to find the eigenvalues of these operators:

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

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

For B:
[tex]tf=\lambda f[/tex]
[tex]\lambda =t [/tex]

The eigenvalues are the variable t?

For B;
[tex]Cf=f=\lambda f[/tex]
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 [tex]<f,g>=\int_0^1 f(t)g(t)dt[/tex]

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

I need to find the eigenvalues of these operators:

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

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:
[tex]tf=\lambda f[/tex]
[tex]\lambda =t [/tex]

The eigenvalues are the variable t?

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

For B;
[tex]Cf=f=\lambda f[/tex]
The eigenvalue is 1.

Confused about B.