A question from the field of Linear Algebra. If I have a linear transformation $T$ that is one-to-one and onto, would that mean that the $T^{-1}$ will also be linear? If so, is there any general proof for it?
Thanks
Asked
Active
Viewed 1.6k times
14
-
3Yes, if "correspondent" means onto. And there is a general proof for it. – Michael Greinecker Jan 31 '13 at 10:18
-
2Where you use the word "opposite", most would use "inverse". – Gerry Myerson Jan 31 '13 at 10:22
-
It is called "the inverse" of a linear transformation. – DonAntonio Jan 31 '13 at 10:24
2 Answers
26
Hint: Write down the equations saying that $T$ is linear. Now replace each variable $x$ with $T^{-1}(y)$ throughout, and apply $T^{-1}$ to both sides of the equation. Simplify.
Marc van Leeuwen
- 115,048
13
Use the relation $$T^{-1} \circ T = {\rm Id}$$ and linearity of $T$ and $\rm Id$ to obtain $$T^{-1} (a T(v) + b T(w)) = av + bw.$$ Now write $v' = T(v)$ and $w' = T(w)$. We get $$T^{-1} (a v' + b w') = aT^{-1}(v') + b T^{-1}(w').$$
Marek
- 6,196