Evaluating T* at the given vector

• December 18th 2010, 03:11 PM
alice8675309
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.
• December 18th 2010, 04:47 PM
Drexel28
Quote:

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?
• December 18th 2010, 04:58 PM
alice8675309
Quote:

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.
• December 19th 2010, 12:20 AM
Opalg
Quote:

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 $(Tg)(t) = g'(t) + 3g(t) = b_1+3(b_0+b_1t)$.
• December 19th 2010, 05:08 AM
alice8675309
Quote:

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.
• December 19th 2010, 06:32 AM
Opalg
Quote:

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 $a_1 = 6$.