# Isomorphism math question

1. May 16, 2005

### laminatedevildoll

Am I doing this right? I'd appreciate any feedback.

Let T:U ---> v be an isomorphism. Show that T^-1: V----> U is linear.

i. T^-1(0) = 0

ii. T^-1(-V) = -T^-1(V)
T^-1(-0) = T^-1(0+0)
= T^-1(0) + T^-1(0)

T^-1(0) = 0
T^-1(-V) = T^-1((-1)V)
=(-1)T^-1(V)
= -T^-1(V)

If T[x,y,z] = [x-y, y-z, x+z]
Then T is one-to-one right?

How do I show that T is onto?

2. May 16, 2005

### mathwonk

what is your definition of isomorphism?

my definition of isomorphism for a linear map T:U-->V is that there exists a linear map S:V-->U such that SoT and ToS are both the identity maps.

then my version of your statement would be to show that a bijective linear map is an isomorphism, i.e. it has an inverse which is linear.

i.e. to answer you question, your definition of isomorphism probably assumes T is onto.

3. May 16, 2005

### laminatedevildoll

An isomorphism is something that is both one-ton-one and onto. But, we were also given that if T: W-->V and S:V-->U are linear transformations, then SoT:W-->U is linear.

4. May 16, 2005

### dextercioby

$\hat{T}$ linear:

$$\hat{T}(ax+by)=a\hat{T}(x)+b\hat{T}(y)$$ (1)

There exists $\hat{T}^{-1}$ so that [itex] \hat{T}^{-1}\left(\hat{T}(x)\right)=\hat{1}x=x [/tex] (2)

Therefore

$$\hat{T}^{-1}\left(\hat{T}\left(ax+by\right)\right)=\hat{1}(ax+by)=ax+by$$ (3) (by virtue of the definition of isomorphism of vector spaces)

From (1),it follows that

$$\hat{T}^{-1}\left(\hat{T}\left(ax+by\right)\right)=T^{-1}\left(a\hat{T}(x)+b\hat{T}(y)\right)$$ (4)

Now,again from (2),i write

$$ax+by=a\hat{T}^{-1}\left(\hat{T}(x)\right)+b\hat{T}^{-1}\left(\hat{T}(y)\right)$$ (5)

Comparing (3),(4) & (5),one gets

$$T^{-1}\left(a\hat{T}(x)+b\hat{T}(y)\right)=a\hat{T}^{-1}\left(\hat{T}(x)\right)+b\hat{T}^{-1}\left(\hat{T}(y)\right)$$ (6)

Q.e.d.

Daniel.

5. May 16, 2005

### mathwonk

by the way, you may not appreciate this for a while, but the word isomorphism should not be defined as your source does, but as i did above. the point is that my definition works for all types of maps, linear, continuous, differentiable, group homomorphism, whatever.

i.e. in many settings a morphism which is one one and onto is still not an isomorphism; e.g. the map taking x to x^3 on the real line does not have a differentiable inverse, but is one one and onto, hence is not a differentiable isomorphism.