A vector space W over the real numbers is the set of all 2 x 2

Click For Summary

Homework Help Overview

The problem involves demonstrating that a specific map T from R4 to a vector space W of 2 x 2 Hermitian matrices is an isomorphism. The discussion centers around the properties required for T to be considered an isomorphism, including linearity, injectivity, and surjectivity.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the necessary properties for T to be an isomorphism, including linearity, injectivity, and surjectivity. There are attempts to clarify the definition of isomorphism and its implications for the algebraic structure of the sets involved. Some participants suggest exploring linear combinations of matrices to demonstrate linearity and bijectivity.

Discussion Status

The discussion is active, with participants providing hints and suggestions for verifying the properties of the map T. There is an emphasis on understanding the algebraic structure of the vector space and ensuring that the operations are preserved. Multiple interpretations of the requirements for isomorphism are being explored.

Contextual Notes

Some participants question the original poster's understanding of isomorphism and the necessary algebraic structures involved. There is a focus on ensuring that the definitions and properties are correctly applied in the context of vector spaces and linear maps.

ZWAN
Messages
1
Reaction score
0

Homework Statement


A vector space W over the real numbers is the set of all 2 x 2 Hermitian matrices. Show that the map T defined as:

T(x,y,z,t) =
[t+x y+iz]
[y-iz t-x]

from R4 to W is an isomorphism.

Homework Equations





The Attempt at a Solution


I know that for the map to be isomorphic it has to have the following properties:
- one to one (injective) and onto (surjective)
- ker(T) = {0} and range(T) = W
- The inverse map T^-1 has to exist
- dimension of both R4 and W has to be the same.

I don't really know how to apply these properties in showing that the map is isomorphic. Any help would be appreciated. Thanks in advanced!
 
Physics news on Phys.org


An isomorphism is by definition a linear bijection, and a bijection is by definition a map that's both injective and surjective. So there are three things you must verify:

1. T is linear.
T(ax+by)=...(use what you know about T here)...aTx+bTy.​
2. T is injective.
Show that Tx=Ty implies x=y.​
3. T is surjective.
Let z be an arbitrary 2×2 hermitian matrix. Prove that there's an x in ℝ4 such that Tx=z.​
 


a hint, sort of:

can you find linearly independent (over R, not C) 2x2 matrices in Mat2x2(C), A,B,C,D such that:

T(x,y,z,t) = xA + yB + zC + tD?

linearity is easy to prove, if so, and so is bijectivity.

(why? because if {A,B,C,D} is LI, it's a basis of something, right?)

another way to prove injectivity:

show ker(T) = {(0,0,0,0)}.

a pre-image of w in W should not be hard to find. figure out what y and z have to be, first, and then tackle x and t. you should get a system of two equations in x and t, easy to solve.
 


1) Your definition of "isomorphism" is incorrect. What you have defined is simply a one-to-one map from one set onto another. You have completely ignored the algebraic structure.

2) And that means your sets must have an algebraic structure. Since you have defined W as a set of matrices, you already have a well defined addition and scalar product. But you are assuming (a, b, c, d)+ (u, v, w, x)= (a+u,b+ v, c+ w, d+ x), and x(a, b, c, d)= (ax, bx, cx, dx) and that really should have been stated.

3) As Deveno said, in addition to showing that the function one to one and onto, you must show that it "preserves" the operations: T(u+ v)= T(u)+ T(v) and T(av)= aT(v).
 

Similar threads

Replies
1
Views
2K
Replies
9
Views
2K
  • · Replies 14 ·
Replies
14
Views
4K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 17 ·
Replies
17
Views
4K
Replies
1
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 24 ·
Replies
24
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K