Proving a linear transformation is an isomorphism

In summary, T is defined as T(a,b) = a + bx and is a linear transformation. To prove it is an isomorphism, we must show that it respects vector addition and scalar multiplication, is one-to-one, and is onto. The operations used in this transformation are addition and multiplication, and to show one-to-one, we can choose values for x to obtain specific values for a and b, ultimately showing that (a,b)=(c,d). Therefore, T is an isomorphism.
  • #1
dmatador
120
1
Define T: F^2 --> P_1(F) by T(a, b) = a + bx (with P_1 denoting P sub 1)

I usually prove problems such as this by constructing a matrix of T using bases for the vector spaces and then proving that the matrix is invertible, but is the following also a viable proof that T is an isomorphism? I know it is not finished, but is it a step in the right direction?

let T(z) = m + nx (T(z) is contained in P_1(F))

so z = (m, n) (z is an element of F^2)

this means that that the general form for all elements in P_1(F) has a pre-image in F^2, which means that T is onto(?), so therefore T is invertible and F^2 is isomorphic to P_1(F).

Is this any start at all? Any suggestions?
 
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  • #2
To prove a function is an isomorphism, you must prove:
1) That it "respects" all operations: that f(a*b)= f(a)*f(b) or any operation *.
2) That is it one-to-one.
3) That it is onto.

Here, the operations are vector addition and scalar multiplication.
T((a,b)+ (c,d))= T((a+c,b+d))= (a+c)+ (b+d)x= a+ bx+ c+ dx= T((a,b))+ T(c,d)
T(\alpha (a, b))= T(\alpha a, \alpha b)= \alpha a+ (\alpha b)x= \alpha(a+ bx)= \alpha T((a,b))

To show "one-to-one", suppose (a,b) and (c,d) were such that T((a,b))= T((c,d)). Then a+ bx= c+ dx for all x[/itex]. What follows from that? (Take x= 0 or x= 1, for example.)

To show "onto", given the linear polynomial u+ vx, does there exist (a,b) such that T((a,b))= u+ vx?
 
  • #3
So in this case you use addition and multiplication, but is that because those are the operations used in the transformation, or because you are simply proving it is a linear transformation?

As for showing that T is one-to-one, is this the right idea?

T((a,b)) = T((c,d))
so then a + bx = c + dx.
then letting x = 0, a = c.
when x = 1, a + b = c + d, and since a = c, we see that b = d.
so (a,b) = (c,d). this means that T is one-to-one.
is this correct?
 

Related to Proving a linear transformation is an isomorphism

What is a linear transformation?

A linear transformation is a mathematical function that maps one vector space to another while preserving the structure of the vector space. This means that it maintains the properties of vector addition and scalar multiplication.

What is an isomorphism?

An isomorphism is a specific type of linear transformation between two vector spaces that is one-to-one and onto. This means that it is both injective (no two inputs map to the same output) and surjective (every element in the output space has at least one corresponding input).

What does it mean to prove a linear transformation is an isomorphism?

To prove a linear transformation is an isomorphism, we must show that it is both one-to-one and onto. This is typically done by showing that the transformation has an inverse that maps the output space back to the input space, and that this inverse also preserves the structure of the vector space.

What are some common methods for proving a linear transformation is an isomorphism?

Some common methods for proving a linear transformation is an isomorphism include showing that the transformation is both injective and surjective, showing that the transformation has a matrix representation with a non-zero determinant, or using the rank-nullity theorem.

Why is proving a linear transformation is an isomorphism important?

Proving that a linear transformation is an isomorphism is important because it allows us to establish an equivalence between two vector spaces. This can be useful in many areas of mathematics and physics, as it allows us to translate problems from one space to another and find solutions more easily.

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