# Inverse linear transformation (think I got it, but don't have answer to compare)

1. Jul 30, 2009

### phyzmatix

1. The problem statement, all variables and given/known data

Define $$T:P_1 \rightarrow P_1$$ by T(a+bx)=(a-b)+ax. Check as to whether T has an inverse or not and if it has, find $$T^{-1}$$

2. Relevant equations

$$T^{-1}(T(v))=v$$

3. The attempt at a solution

The range of T is P1, so T is one-to-one. The inverse of T is therefore

$$T^{-1}((a-b)+ax)=(a+bx)$$

Is my reasoning sound?

Thanks for the help!
phyz

Last edited: Jul 30, 2009
2. Jul 30, 2009

### n!kofeyn

Yes, T is a linear transformation from P1 to P1, so if it is injective or surjective it is invertible. Should you show T is one-to-one using the definition or by showing null(T)={0}?

Your T-1 needs to be actually defined. As defined, T's range is P2. So let c+dx+ex2 be in P2. Define T-1:P2→P1 by T-1(c+dx+ex2)=d+(d-c)x. Now check that T-1 is actually the inverse of T by showing T-1(T(a+bx))=a+bx.

3. Jul 30, 2009

### phyzmatix

Hi there...sorry, but I made a typo

Should be $$T:P_1 \rightarrow P_1$$ not $$T:P_1 \rightarrow P_2$$. Fixed it now though. They don't say how you should determine whether T has an inverse or not.

4. Jul 30, 2009

### n!kofeyn

Oh okay, no problem. That makes more sense, as that confused me a little bit. Then just remove the ex2 from everywhere it appears above.

Read http://books.google.com/books?id=BNsOE3Gp_hEC&lpg=PP1&dq=sheldon axler linear algebra&pg=PA53" for invertibility of linear maps. Look at Theorem 3.21. To show T is invertible, all you need to do is show that it is injective or onto since it is a linear map from P1 to P1. I would recommend showing that if T(a+bx)=T(c+dx), then a+bx=c+dx to show it is one-to-one. Then just define the inverse of T as above, and you're done.

Last edited by a moderator: Apr 24, 2017
5. Jul 31, 2009

### phyzmatix

Thanks for that! Have a great weekend!

6. Jul 31, 2009

### n!kofeyn

Haha. No problem, and the same to you!