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

[phyzmatix]

Homework Statement

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}

Homework Equations

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

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

 

[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.

 

[phyzmatix]

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.

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.

 

[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.

 

[phyzmatix]

Thanks for that! Have a great weekend!

 

[n!kofeyn]

Haha. No problem, and the same to you!