# Thread: Evaluating T* at the given vector

1. ## Evaluating T* at the given vector

Hey guys, I'm sorry for all the posts lately but i'm studying for finals so just bare with me. However, Im trying to follow an example in my book that is similar to one of my practice problems but I can't seem to see where one part is coming from.

For the following inner product space V and linear operator T on V, evaluate T* at the given vector in V:

V=P_1(R) with <f,g> $\int_{-1}^{1}$f(t)g(t)dt, T(f)=f'+3f, f(t)=4-2t

Solution:

Let g(t)=b_0+b_1t, where b_0 and b_1 are arbitrarily real numbers. Then (Tg)(t)=(3b_0+b_1+3b_1t) so we must match $\int_{-1}^{1}$(T*f)(t)(b_0+b_1t)dt= $\int_{-1}^{1}$(4-2t)(3b_0+b_1+3b_1t)dt=24b_0+4b_1.

Since (T*f)(t)=a_0+a_1t, b_0=1, b_1=0 and b_0=0, b_1=1. This gives (T*f)(t)=12+6t

The part that I'm confused with is how did they get (Tg)(t)=(3b_0+b_1+3b_1t)? Other than I follow and understand.

2. Originally Posted by alice8675309
Hey guys, I'm sorry for all the posts lately but i'm studying for finals so just bare with me. However, Im trying to follow an example in my book that is similar to one of my practice problems but I can't seem to see where one part is coming from.

For the following inner product space V and linear operator T on V, evaluate T* at the given vector in V:

V=P_1(R) with <f,g> $\int_{-1}^{1}$f(t)g(t)dt, T(f)=f'+3f, f(t)=4-2t

Solution:

Let g(t)=b_0+b_1t, where b_0 and b_1 are arbitrarily real numbers. Then (Tg)(t)=(3b_0+b_1+3b_1t) so we must match $\int_{-1}^{1}$(T*f)(t)(b_0+b_1t)dt= $\int_{-1}^{1}$(4-2t)(3b_0+b_1+3b_1t)dt=24b_0+4b_1.

Since (T*f)(t)=a_0+a_1t, b_0=1, b_1=0 and b_0=0, b_1=1. This gives (T*f)(t)=12+6t

The part that I'm confused with is how did they get (Tg)(t)=(3b_0+b_1+3b_1t)? Other than I follow and understand.
What is $T^\ast$? Some kind of dual homomorphism?

3. Originally Posted by Drexel28
What is $T^\ast$? Some kind of dual homomorphism?
Sorry about that, no, T* is the adjoint of T. Sorry for any confusion.

4. Originally Posted by alice8675309
Hey guys, I'm sorry for all the posts lately but i'm studying for finals so just bare with me. However, Im trying to follow an example in my book that is similar to one of my practice problems but I can't seem to see where one part is coming from.

For the following inner product space V and linear operator T on V, evaluate T* at the given vector in V:

V=P_1(R) with <f,g> $\int_{-1}^{1}$f(t)g(t)dt, T(f)=f'+3f, f(t)=4-2t

Solution:

Let g(t)=b_0+b_1t, where b_0 and b_1 are arbitrarily real numbers. Then (Tg)(t)=(3b_0+b_1+3b_1t) so we must match $\int_{-1}^{1}$(T*f)(t)(b_0+b_1t)dt= $\int_{-1}^{1}$(4-2t)(3b_0+b_1+3b_1t)dt=24b_0+4b_1.

Since (T*f)(t)=a_0+a_1t, b_0=1, b_1=0 and b_0=0, b_1=1. This gives (T*f)(t)=12+6t

The part that I'm confused with is how did they get (Tg)(t)=(3b_0+b_1+3b_1t)? Other than I follow and understand.
You are told that $T(f) = f'+3f$, so $T(g) = g'+3g$. If $g(t) = b_0+b_1t$ then $g'(t) = b_1$. Hence $(Tg)(t) = g'(t) + 3g(t) = b_1+3(b_0+b_1t)$.

5. Originally Posted by Opalg
You are told that $T(f) = f'+3f$, so $T(g) = g'+3g$. If $g(t) = b_0+b_1t$ then $g'(t) = b_1$. Hence $(Tg)(t) = g'(t) + 3g(t) = b_1+3(b_0+b_1t)$.
Thanks! But after finally getting passed that part it seems I'm confused at how they got (T*f)(t)=12+6t
I understand how the right integral id 24b_0+4b_1..but from there it kind of jumps and my book is not helpful at all.

6. Originally Posted by alice8675309
Thanks! But after finally getting passed that part it seems I'm confused at how they got (T*f)(t)=12+6t
I understand how the right integral id 24b_0+4b_1..but from there it kind of jumps and my book is not helpful at all.
Let $(T^*f)(t) = a_0+a_1t$. Then

$\displaystyle\langle T^*f,g\rangle = \int_{-1}^1\!\!\!(a_0+a_1t)(b_0+b_1t)\,dt = \Bigl[a_0b_0t + \tfrac12(a_1b_0+a_0b_1)t^2 + \tfrac13a_1b_1t^3\Bigr]_{-1}^1 = 2a_0b_0 + \tfrac23a_1b_1.$

You want that to be equal to $24b_0+4b_1$. So compare coefficients of $b_0$ and $b_1$ to get $a_0=12$ and $a_1 = 6$.