Prove that this is a linear transformation

  • #1
162
0
The problem statement has been attached.

To show that T : V →R is a linear function
It must satisfy 2 conditions:
1) T(cv) = cT(v) where c is a constant
and
2) T(u+v) = T(u)+T(v)


For condition 1)
T(cv)=∫cvdx from 0 to 1 (I don't know how to put limits into the integral.
cT(v)=c∫vdx (pulling a constant out of an integral)
cT(v)=cT(v)

I think I did the work for condition 1 right?


For condition 2)
T(u+v) where u=u1 and v=v1
T(u+v)=∫(u+v)dx
T(u+v)=∫(u)dx + ∫(v)dx (this is called the sum rule of integration, I think).
T(u+v)=T(u)+T(v)

Did i do this right?
 

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Answers and Replies

  • #2
35,054
6,791
The problem statement has been attached.

To show that T : V →R is a linear function
It must satisfy 2 conditions:
1) T(cv) = cT(v) where c is a constant
and
2) T(u+v) = T(u)+T(v)


For condition 1)
T(cv)=∫cvdx from 0 to 1 (I don't know how to put limits into the integral.
cT(v)=c∫vdx (pulling a constant out of an integral)
cT(v)=cT(v)

I think I did the work for condition 1 right?


For condition 2)
T(u+v) where u=u1 and v=v1
T(u+v)=∫(u+v)dx
T(u+v)=∫(u)dx + ∫(v)dx (this is called the sum rule of integration, I think).
T(u+v)=T(u)+T(v)

Did i do this right?

The 2nd part looks OK, but both parts could be cleaned up some.

For the first part, start with T(cv) and keep working until you get to cT(v), connecting each pair of successive equal expressions with '='. In your second line you start off with cT(v), which is you're trying to show is equal to T(cv).

The work should look like this:
T(cv) = ##\int_0^1 cv~dx = c\int_0^1 v~dx## = cT(v).

In the second part, you have u = u1 and v = v1, but then you never use them, so why are they there?

The second part should look like this: T(u + v) = ##\int_0^1 u +v~dx## = ... = T(u) + T(v)

I used LaTeX for the integrals. Right-click any of the integrals to see how I did it.
 
  • #3
162
0
Is there any point in providing the limits of integration? I don't see the use except to define the function as continous on that interval.
 

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