Algebraic Structures | Question-2
Solution
Recall the definition of \( B \):
From Question 1, we established that:
Characterization of \( B \):
A matrix \( X = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) belongs to \( B \) if and only if:
\[ c = 0 \quad \text{and} \quad d = a \]
Thus, matrices in \( B \) have the form:
\[ X = \begin{pmatrix} a & b \\ 0 & a \end{pmatrix}, \quad a,b \in \mathbb{R} \]
Express \( P \) and \( Q \) in general form:
Since \( P, Q \in B \), we can write them as:
General Forms:
\[ P = \begin{pmatrix} p_1 & p_2 \\ 0 & p_1 \end{pmatrix}, \quad \text{where } p_1, p_2 \in \mathbb{R} \]
\[ Q = \begin{pmatrix} q_1 & q_2 \\ 0 & q_1 \end{pmatrix}, \quad \text{where } q_1, q_2 \in \mathbb{R} \]
Compute \( P + Q \):
Matrix Addition:
\[ P + Q = \begin{pmatrix} p_1 & p_2 \\ 0 & p_1 \end{pmatrix} + \begin{pmatrix} q_1 & q_2 \\ 0 & q_1 \end{pmatrix} = \begin{pmatrix} p_1 + q_1 & p_2 + q_2 \\ 0 + 0 & p_1 + q_1 \end{pmatrix} \]
\[ P + Q = \begin{pmatrix} p_1 + q_1 & p_2 + q_2 \\ 0 & p_1 + q_1 \end{pmatrix} \]
Verify that \( P + Q \) satisfies the conditions for \( B \):
Check Conditions:
Let \( R = P + Q = \begin{pmatrix} r_1 & r_2 \\ r_3 & r_4 \end{pmatrix} \), where:
\[ r_1 = p_1 + q_1, \quad r_2 = p_2 + q_2, \quad r_3 = 0, \quad r_4 = p_1 + q_1 \]
Condition 1 (Lower-left entry): \( r_3 = 0 \) ✅
Condition 2 (Diagonal equality): \( r_4 = p_1 + q_1 = r_1 \) ✅
Thus, \( R = P + Q \) has the required form:
\[ R = \begin{pmatrix} r_1 & r_2 \\ 0 & r_1 \end{pmatrix} \]
where \( r_1 = p_1 + q_1 \) and \( r_2 = p_2 + q_2 \) are real numbers.
Alternative proof using the definition directly:
Direct Proof Using Commutativity:
Since \( P, Q \in B \), we know:
\[ MP = PM \quad \text{and} \quad MQ = QM \]
Now consider \( M(P + Q) \):
\[ M(P + Q) = MP + MQ \quad \text{(by distributivity of matrix multiplication)} \]
\[ = PM + QM \quad \text{(since \( MP = PM \) and \( MQ = QM \))} \]
\[ = (P + Q)M \quad \text{(by distributivity of matrix multiplication)} \]
Thus, \( M(P + Q) = (P + Q)M \), which means \( P + Q \) commutes with \( M \).
Therefore, by definition of \( B \), \( P + Q \in B \).
Conclusion
We have proven that \( B \) is closed under matrix addition:
Final Answer: ✅ PROVED
We have shown using two different approaches:
- Structural Approach: Using the explicit form of matrices in \( B \), we showed that if \( P = \begin{pmatrix} p_1 & p_2 \\ 0 & p_1 \end{pmatrix} \) and \( Q = \begin{pmatrix} q_1 & q_2 \\ 0 & q_1 \end{pmatrix} \), then \( P + Q = \begin{pmatrix} p_1 + q_1 & p_2 + q_2 \\ 0 & p_1 + q_1 \end{pmatrix} \) which has the required form.
- Direct Proof Approach: Using the commutativity property and distributivity of matrix multiplication, we showed that \( M(P + Q) = (P + Q)M \), proving \( P + Q \in B \).
Important Note: This result shows that \( B \) is closed under addition, which is one of the properties needed for \( B \) to be a subspace of \( A \).
Summary: ✅ We have successfully proven that if \( P, Q \in B \), then \( P + Q \in B \).