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Consider the group homomorphism phi:M2(R)--->R,given by phi(A)=trc(A). The ker of phi is isomorphic to which of the map?
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Consider the group homomorphism phi:M2(R)--->R,given by phi(A)=trc(A)....
Group Homomorphism from M2(R) to R

The given group homomorphism is φ : M2(R) → R defined by φ(A) = trc(A), where M2(R) denotes the set of 2x2 matrices with real entries, and trc(A) denotes the trace of the matrix A.

Definition of Group Homomorphism

A group homomorphism is a function between two groups that preserves the group operation. In this case, the groups are M2(R) and R.

Kernel of the Homomorphism

The kernel of a homomorphism is the set of elements in the domain that map to the identity element in the codomain. In this case, we need to find the set of matrices A in M2(R) such that φ(A) = 0.

Finding the Kernel

Let A = [[a, b], [c, d]] be a matrix in M2(R). Then the trace of A is trc(A) = a + d.

To find the kernel, we need to solve the equation φ(A) = trc(A) = 0.

a + d = 0

This equation implies that a = -d.

So, the kernel consists of matrices of the form A = [[a, b], [-a, d]], where a and d are real numbers.

Isomorphism and the Kernel

To determine which map the kernel is isomorphic to, we need to consider the properties of the kernel.

Kernel of φ is isomorphic to the additive group of real numbers (R, +)

The kernel of φ consists of matrices of the form A = [[a, b], [-a, d]], where a and d are real numbers. This set forms an additive group isomorphic to the group of real numbers (R, +) under addition.

Explanation

The isomorphism between the kernel of φ and the additive group of real numbers (R, +) can be established by considering the following:

1. Identity Element: The zero matrix [[0, 0], [0, 0]] is the identity element in the kernel, which corresponds to the additive identity in the group of real numbers (R, +), i.e., 0.

2. Closure: Addition of matrices in the kernel results in another matrix in the kernel, and addition of real numbers in (R, +) results in another real number.

3. Inverses: Every matrix in the kernel has an inverse within the kernel, where the additive inverse of [[a, b], [-a, d]] is [[-a, -b], [a, -d]]. Similarly, every real number in (R, +) has an additive inverse.

4. Associativity and Commutativity: Addition of matrices in the kernel and addition of real numbers in (R, +) satisfy associativity and commutativity.

Therefore, the kernel of φ is isomorphic to the additive group of real numbers (R, +).

Summary

The kernel of the group homomorphism φ : M2(R) → R, given by φ(A) = trc(A), is isomorphic to the additive group of real numbers (R, +). This is because the elements in the kernel, matrices of the form A =
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Consider the group homomorphism phi:M2(R)--->R,given by phi(A)=trc(A)....
Group homomorphism phi:M2(R)--->R

The group homomorphism phi:M2(R)--->R is defined by phi(A)=trc(A), where M2(R) is the group of 2x2 matrices with real entries, R is the group of real numbers, and trc(A) denotes the trace of matrix A.

Definition of Trace

The trace of a matrix A = [a11 a12; a21 a22] is defined as the sum of its diagonal elements, i.e., trc(A) = a11 + a22.

Group Homomorphism

A group homomorphism is a map between two groups that preserves the group structure. In this case, phi is a group homomorphism from M2(R) to R because it satisfies the following properties:

1. phi(A + B) = trc(A + B) = trc(A) + trc(B) for all A, B in M2(R)
2. phi(kA) = trc(kA) = k * trc(A) for all A in M2(R) and k in R

Ker(phi)

The kernel of a group homomorphism phi is the set of elements in the domain that map to the identity element in the codomain. In this case, we need to find the matrices A in M2(R) such that phi(A) = 0.

Let's consider a matrix A = [a11 a12; a21 a22]. The trace of A is given by trc(A) = a11 + a22. For phi(A) to be 0, we have the equation a11 + a22 = 0.

Isomorphism with R2

To find the kernel of phi, we need to solve the equation a11 + a22 = 0. This equation represents a line in the plane of 2x2 matrices. The kernel of phi corresponds to the set of matrices that lie on this line.

Considering the equation a11 + a22 = 0, we can rewrite it as a11 = -a22. This means that for any value of a22, we can choose a11 = -a22 to satisfy the equation. Therefore, the kernel of phi is isomorphic to R, the set of real numbers.

We can establish an isomorphism between the kernel of phi and R by mapping each real number r to the matrix A = [-r r; 0 0]. The trace of A is trc(A) = -r + 0 = -r, which corresponds to the real number r. This mapping is a bijection and preserves the group structure, so the kernel of phi is isomorphic to R.

Summary

- The group homomorphism phi:M2(R) ---> R is defined by phi(A) = trc(A), where trc(A) denotes the trace of matrix A.
- The kernel of phi is the set of matrices A in M2(R) such that trc(A) = 0.
- The kernel of phi is isomorphic to R, the set of real numbers.
- An isomorphism can be established by mapping each real number r to the matrix A = [-r r; 0 0], which preserves the group structure and the trace.
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Consider the group homomorphism phi:M2(R)--->R,given by phi(A)=trc(A). The ker of phi is isomorphic to which of the map?
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Consider the group homomorphism phi:M2(R)--->R,given by phi(A)=trc(A). The ker of phi is isomorphic to which of the map? for Mathematics 2024 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about Consider the group homomorphism phi:M2(R)--->R,given by phi(A)=trc(A). The ker of phi is isomorphic to which of the map? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider the group homomorphism phi:M2(R)--->R,given by phi(A)=trc(A). The ker of phi is isomorphic to which of the map?.
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