Is T an Isomorphism and How Can It Be Proven?

[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?

 

[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.

 

[laminatedevildoll]

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.

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.

 

[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)<br /> <br /> Therefore<br /> <br /> $$\hat{T}^{-1}\left(\hat{T}\left(ax+by\right)\right)=\hat{1}(ax+by)=ax+by$$ (3) (by virtue of the definition of isomorphism of vector spaces)<br /> <br /> From (1),it follows that<br /> <br /> $$\hat{T}^{-1}\left(\hat{T}\left(ax+by\right)\right)=T^{-1}\left(a\hat{T}(x)+b\hat{T}(y)\right)$$ (4)<br /> <br /> Now,again from (2),i write<br /> <br /> $$ax+by=a\hat{T}^{-1}\left(\hat{T}(x)\right)+b\hat{T}^{-1}\left(\hat{T}(y)\right)$$ (5)<br /> <br /> Comparing (3),(4) & (5),one gets<br /> <br /> $$T^{-1}\left(a\hat{T}(x)+b\hat{T}(y)\right)=a\hat{T}^{-1}\left(\hat{T}(x)\right)+b\hat{T}^{-1}\left(\hat{T}(y)\right)$$ (6)<br /> <br /> Q.e.d.<br /> <br /> Daniel.[/itex]

 

[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.