# Linear Algebra: Linear transformation and Isomorphism

1. Mar 12, 2009

### MathNoob22

1. The problem statement, all variables and given/known data
Let T: V $$\rightarrow$$ Z be a linear transformation of a vector space V onto a vector space Z. Define the mapping

$$\bar{T}$$: V/N(T) $$\rightarrow$$ Z by $$\bar{T}$$(v + N(T)) = T(v)

for any coset v+N(T) in V/N(T).

a) Prove that $$\bar{T}$$ is well-defined; that is, prove that if v+N(T)=v'+N(T), then T(v)=T(v').

b) Prove that $$\bar{T}$$ is linear.

c) Prove that $$\bar{T}$$ is an isomorphism.

2. Relevant equations
Proving part (a) will prove that the transformation is a one-to-one transformation. This is important for part (c).

In order to prove part (b), use the fact that for T: V$$\rightarrow$$W, T is linear if and only if T(cx + y) = cT(x)+T(y) for all x,y $$\in$$ V and c $$\in$$ F.

To prove that a transformation is an isomorphism, part (c), there must exist a linear transformation $$\bar{T}$$ that is invertible ($$\bar{T}$$$$^{-1}$$). $$\bar{T}$$$$^{-1}$$ exists iff $$\bar{T}$$ is one-to-one and onto.

3. The attempt at a solution
(a) If v+N(T) = v'+N(T), then v+N(T)-N(T) = v'+N(T)-N(T). Thus v = v'. The vectors v and v' are the same without the nullspace of T.
If v=v', then T(v) = T(v'). This also means it is one-to-one.

I'm not sure if what I have is a valid proof for part (a). The part that I'm not certain about is if I am actually able to subtract the nullspace of T from both sides.

(b) If $$\bar{T}$$ = T(v), let cx+y = v+N(T). Then $$\bar{T}$$(cx+y) = T(cx+y-N(T)) = cT(x)+T(y)-T(N(T)). This is the first part of the proof.

I am having trouble with the next part of the proof. I have to prove that c$$\bar{T}$$(x) + $$\bar{T}$$(y) = cT(x)+T(y)-T(N(T)) (the right hand side is from the first part of the proof), but I'm not sure how to plug x and y into $$\bar{T}$$.

(c) A transformation is an isomorphism if it is one-to-one and onto.
To prove one-to-one: Since v+N(T)=v'+N(T), then v=v'. If v=v', then T(v)=T(v'). T(v)=T(v') implies that v=v', this is the definition of one-to-one.

I'm having trouble proving that $$\bar{T}$$ is onto. From the question, we know that T is a linear transformation that is onto, but how do I relate it to $$\bar{T}$$?

Any and all help is much appreciated!

2. Mar 12, 2009

### Focus

Your part a is incorrect. You can't take away N(T) from both sides, its a set. You could have that v+n=v'+m for m not equal to n.

For b) you really only need to show that $\overline{T}(cx+ky+N(T))=c\overline{T}(x+N(T))+k\overline{T}(y+N(T))$ which uses the linearity of T.

c) you are making the same mistake again. Plus to show its injective you need to take $\overline{T}(x+N(T))=\overline{T}(y+N(T))$ and show that $x+N(T)=y+N(T)$.

I hope some of that made sense. Been a long day.

3. Mar 12, 2009

### yyat

N(T)-N(T) is not 0, it is N(T), check the definition of the quotient vector space. It is in general not true that v=v', but you can still conclude that T(v)=T(v') by using the definition of N(T).

No, one-to-one means that if T(v)=T(v') then v=v', it is the other way around. Besides, if T were 1-to-1, then N(T) would contain only the 0 vector.

You have to show that $$\bar{T}$$(cx+y+N(T))=c$$\bar{T}$$(x+N(T))+$$\bar{T}$$(y+N(T)). Remember that you should only apply $$\bar{T}$$ to cosets, not the original vectors. To prove this part, use that $$\bar{T}$$(v+N(T))=T(v) for any v and linearity of T.

You cannot conclude that v=v', same reason as above.

That T is onto implies that for any v in Z you can find a w in V with T(w)=v. Now you want to find a coset u+N(T) such that $$\bar{T}$$(u+N(T))=v. What would be a good choice for u?

4. Mar 15, 2009

### MathNoob22

Hello again. Thanks for all the help and advice. This is the new proof I have now.

a) Prove that $$\bar{T}$$ is well defined; that is prove that if v+N(T)=v'+N(T), then T(v)=T(v').

By definition, N(T) is the set of all vectors x in V such that T(x)=0.
If v+N(T)=v'+N(T), both cosets have vectors v and v' added to the same Nullspace of T.
v+N(T) = v'+N(T)
$$\bar{T}$$(v+N(T)) = $$\bar{T}$$(v'+N(T))
T(v)= T(v')

Is this enough for the first proof? I took the $$\bar{T}$$ transformation of both v+N(T) and v'+N(T). Since they were equal, the $$\bar{T}$$ transformation should also be equal.

b) Prove that $$\bar{T}$$ is linear.

$$\bar{T}$$(cx+y+N(T)) = $$\bar{T}$$(cx+N(T)) + $$\bar{T}$$(y+N(T))
$$\bar{T}$$(cx+y+N(T))
= T(cx+y)
= T(cx) + T(y)
= cT(x) + T(y)

$$\bar{T}$$(cx+N(T)) + $$\bar{T}$$(y+N(T))
= T(cx) + T(y)
= cT(x) + T(y)
cT(x) + T(y) = cT(x) + T(y)

I think my work for part (b) is enough to prove that $$\bar{T}$$ is linear. Is there anything else I need?

c) Prove that $$\bar{T}$$ is an isomorphism.

$$\bar{T}$$ is an isomorphism iff $$\bar{T}$$ is one-to-one and onto.
One-to-One: Let T(x)=T(y) for any x,y $$\in$$ V.
$$\bar{T}$$(x+N(T)) = $$\bar{T}$$(y+N(T))
If the transformation of the coset of x+N(T) is equal to the transformation of the coset of y+N(T), the coset x+N(T) must be equal to the coset of y+N(T). As stated in part (a), N(T) is the same in both cosets (because the Nullspace does not change). So, x=y. If T(x)=T(y) implies x=y, then $$\bar{T}$$ must be one-to-one.

Onto: Let $$\bar{T}$$(x+N(T)) = T(x) for any x $$\in$$ V/N(T).
$$\bar{T}$$ is onto if there exists a vector y in Z such that T(x) = y for any y $$\in$$ Z.
So, $$\bar{T}$$(u+N(T)) = T(u) = y
y = T(x)
Then, T(u) = T(x)
$$\bar{T}$$(u + N(T)) = $$\bar{T}$$(x + N(T))
u = x
Therefore, for any vector y $$\in$$ Z, there exists a vector x $$\in$$ V; this means it is onto.

Since it is both one-to-one and onto, $$\bar{T}$$ is an isomorphism.

I'm not sure if the proof I've given is correct or if it is adequate enough. That is my biggest worry that I do not provide enough work.