Math Problem Statement

Let G be the group of real 2x2 matrices under matrix addition. Let H be the subset of G of all matrices of the form [[a, b], [c, d]] for which a + b = c + d. Prove that H is a subgroup of G.

Solution

The question asks us to prove that HGH \leq G, where:

  1. GG is the group of all 2×22 \times 2 matrices with real entries under matrix addition.
  2. HH is the subset of GG consisting of matrices of the form: (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix} where a+b=c+da + b = c + d.

To show that HGH \leq G, we need to prove that HH is a subgroup of GG. We will do this by checking the subgroup criteria:

  1. Closure under addition: If A,BHA, B \in H, then A+BHA + B \in H.
  2. Existence of an identity element: The zero matrix should be in HH.
  3. Existence of inverses: For each AHA \in H, the additive inverse A-A should also be in HH.

Let's go through each condition.

Step 1: Closure under addition

Suppose A=(abcd)A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} and B=(efgh)B = \begin{pmatrix} e & f \\ g & h \end{pmatrix} are elements of HH. This means: a+b=c+dande+f=g+h.a + b = c + d \quad \text{and} \quad e + f = g + h. Now, consider A+B=(a+eb+fc+gd+h)A + B = \begin{pmatrix} a+e & b+f \\ c+g & d+h \end{pmatrix}.

For A+BA + B to be in HH, it must satisfy (a+e)+(b+f)=(c+g)+(d+h)(a + e) + (b + f) = (c + g) + (d + h). Notice that: (a+e)+(b+f)=(a+b)+(e+f)=(c+d)+(g+h)=(c+g)+(d+h).(a + e) + (b + f) = (a + b) + (e + f) = (c + d) + (g + h) = (c + g) + (d + h). Thus, A+BHA + B \in H, showing closure under addition.

Step 2: Identity element

The identity element in GG under addition is the zero matrix O=(0000)O = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}. Since 0+0=0+00 + 0 = 0 + 0, the zero matrix satisfies a+b=c+da + b = c + d and therefore belongs to HH.

Step 3: Existence of inverses

For A=(abcd)HA = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in H, consider its additive inverse A=(abcd)-A = \begin{pmatrix} -a & -b \\ -c & -d \end{pmatrix}. Since a+b=c+da + b = c + d, we have: (a)+(b)=(a+b)=(c+d)=(c)+(d).(-a) + (-b) = -(a + b) = -(c + d) = (-c) + (-d). Thus, AH-A \in H, so every element in HH has an additive inverse in HH.

Conclusion

Since HH is closed under addition, contains the identity element, and contains inverses, HH is a subgroup of GG. Therefore, we have shown that HGH \leq G.


Would you like further clarification on any step?

Related Questions

  1. What is the difference between a group and a subgroup?
  2. How can we verify the subgroup criteria in general for matrix groups?
  3. What are other examples of conditions that can define subgroups of matrix groups?
  4. Can the same method be applied to prove subgroup properties in groups other than matrices?
  5. What is the significance of additive and multiplicative subgroups in linear algebra?

Tip

Remember, to prove a subset is a subgroup, verifying closure, identity, and inverses with respect to the group's operation is key.

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Math Problem Analysis

Mathematical Concepts

Abstract Algebra
Matrix Groups
Subgroup Criterion

Formulas

Matrix addition
Subgroup criteria

Theorems

Subgroup Test

Suitable Grade Level

College Level (Abstract Algebra)