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The question asks us to prove that \( H \leq G \), where:

1. \( G \) is the group of all \( 2 \times 2 \) matrices with real entries under matrix addition.
2. \( H \) is the subset of \( G \) consisting of matrices of the form:
   \[
   \begin{pmatrix} a & b \\ c & d \end{pmatrix}
   \]
   where \( a + b = c + d \).

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

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

Let's go through each condition.

### Step 1: Closure under addition
Suppose \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) and \( B = \begin{pmatrix} e & f \\ g & h \end{pmatrix} \) are elements of \( H \). This means:
\[
a + b = c + d \quad \text{and} \quad e + f = g + h.
\]
Now, consider \( A + B = \begin{pmatrix} a+e & b+f \\ c+g & d+h \end{pmatrix} \).

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

### Step 2: Identity element
The identity element in \( G \) under addition is the zero matrix \( O = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \). Since \( 0 + 0 = 0 + 0 \), the zero matrix satisfies \( a + b = c + d \) and therefore belongs to \( H \).

### Step 3: Existence of inverses
For \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in H \), consider its additive inverse \( -A = \begin{pmatrix} -a & -b \\ -c & -d \end{pmatrix} \). Since \( a + b = c + d \), we have:
\[
(-a) + (-b) = -(a + b) = -(c + d) = (-c) + (-d).
\]
Thus, \( -A \in H \), so every element in \( H \) has an additive inverse in \( H \).

### Conclusion
Since \( H \) is closed under addition, contains the identity element, and contains inverses, \( H \) is a subgroup of \( G \). Therefore, we have shown that \( H \leq G \).

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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.

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.

In this problem, we prove that a subset H of the group G, consisting of all 2x2 matrices where a + b = c + d, forms a subgroup under matrix addition. This solution uses subgroup criteria, focusing on closure, identity, and inverse properties.

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