Let's solve each problem step by step with explanations. --- ### **51. Let \(A\) and \(B\) be \(n \times n\) matrices. Verify that \(\mathcal{N}(A) \cap \mathcal{N}(B) \subseteq \mathcal{N}(A+B)\).** #### **Step 1: Understand the notation** - \(\mathcal{N}(A)\) is the null space (kernel) of \(A\): the set of all vectors \(x\) such that \(A x = 0\). - \(\mathcal{N}(B)\) is the null space of \(B\). - \(\mathcal{N}(A) \cap \mathcal{N}(B)\) is the set of all \(x\) such that both \(A x = 0\) and \(B x = 0\). - \(\mathcal{N}(A+B)\) is the set of all \(x\) such that \((A+B)x = 0\). #### **Step 2: Show the subset relation** Let \(x \in \mathcal{N}(A) \cap \mathcal{N}(B)\). - Then \(A x = 0\) and \(B x = 0\). Now, consider \((A+B)x\): \[ (A+B)x = Ax + Bx = 0 + 0 = 0 \] So, \(x \in \mathcal{N}(A+B)\). #### **Step 3: Conclusion** Thus, every \(x \in \mathcal{N}(A) \cap \mathcal{N}(B)\) is also in \(\mathcal{N}(A+B)\), so: \[ \boxed{\mathcal{N}(A) \cap \mathcal{N}(B) \subseteq \mathcal{N}(A+B)} \] --- ### **52. Let \(A\) be an \(m \times r\) matrix and \(B\) an \(r \times n\) matrix.** #### **(a) Show that \(\mathcal{N}(B) \subseteq \mathcal{N}(AB)\).** - \(\mathcal{N}(B)\): Set of all \(x\) such that \(B x = 0\). - \(AB\) is an \(m \times n\) matrix. Let \(x \in \mathcal{N}(B)\). Then \(B x = 0\). Now, consider \(AB x\): \[ AB x = A (B x) = A(0) = 0 \] So, \(x \in \mathcal{N}(AB)\). \[ \boxed{\mathcal{N}(B) \subseteq \mathcal{N}(AB)} \] --- #### **(b) Show that \(\mathcal{R}(AB) \subseteq \mathcal{R}(A)\).** - \(\mathcal{R}(AB)\): Range (column space) of \(AB\). - \(\mathcal{R}(A)\): Range of \(A\). Let \(y \in \mathcal{R}(AB)\). Then there exists some \(x\) such that \(y = ABx\). But \(ABx = A(Bx)\). Let \(z = Bx\), so \(y = A z\). Thus, \(y\) is in the range of \(A\). \[ \boxed{\mathcal{R}(AB) \subseteq \mathcal{R}(A)} \] --- ## **Final Answers** **51.** \(\mathcal{N}(A) \cap \mathcal{N}(B) \subseteq \mathcal{N}(A+B)\) **52.** (a) \(\mathcal{N}(B) \subseteq \mathcal{N}(AB)\) (b) \(\mathcal{R}(AB) \subseteq \mathcal{R}(A)\)

Certainly! Let's go through each part in detail, including step-by-step explanations and calculations where relevant. --- ## **Problem 51** **Claim:** For square matrices \(A, B \in \mathbb{R}^{n \times n}\), show that: \[ \mathcal{N}(A) \cap \mathcal{N}(B) \subseteq \mathcal{N}(A + B) \] --- ### **Step-by-step Solution:** ### **Step 1: Recall definitions** - \(\mathcal{N}(A) = \{ x \in \mathbb{R}^n \mid A x = 0 \}\) - \(\mathcal{N}(B) = \{ x \in \mathbb{R}^n \mid B x = 0 \}\) - \(\mathcal{N}(A + B) = \{ x \in \mathbb{R}^n \mid (A + B) x = 0 \}\) --- ### **Step 2: Take an arbitrary vector in the intersection** Let \(x \in \mathcal{N}(A) \cap \mathcal{N}(B)\). This means: \[ A x = 0 \quad \text{and} \quad B x = 0 \] --- ### **Step 3: Check whether \(x\) belongs to \(\mathcal{N}(A + B)\)** Calculate: \[ (A + B) x = A x + B x \] Since \(A x = 0\) and \(B x = 0\): \[ (A + B) x = 0 + 0 = 0 \] Thus, \(x \in \mathcal{N}(A + B)\). --- ### **Step 4: Conclusion** Because our choice of \(x\) was arbitrary in \(\mathcal{N}(A) \cap \mathcal{N}(B)\), and we showed that such \(x\) is in \(\mathcal{N}(A + B)\), it follows that: \[ \boxed{\mathcal{N}(A) \cap \mathcal{N}(B) \subseteq \mathcal{N}(A + B)} \] --- ## **Problem 52** Given matrices: - \(A\) of size \(m \times r\), - \(B\) of size \(r \times n\). --- ### **Part (a): Show that \(\mathcal{N}(B) \subseteq \mathcal{N}(AB)\).** --- ### **Step 1: Understand the null spaces** - \(\mathcal{N}(B) = \{ x \in \mathbb{R}^n \mid B x = 0 \}\) - \(A B\) is an \(m \times n\) matrix, so: \[ \mathcal{N}(A B) = \{ x \in \mathbb{R}^n \mid (A B) x = 0 \} \] --- ### **Step 2: Take an arbitrary vector in \(\mathcal{N}(B)\)** Let \(x \in \mathcal{N}(B)\): \[ B x = 0 \] --- ### **Step 3: Compute \(A B x\)** \[ A B x = A (B x) = A (0) = 0 \] Thus, \(x \in \mathcal{N}(A B)\). --- ### **Step 4: Conclude the subset relation** Since the choice of \(x\) was arbitrary in \(\mathcal{N}(B)\), we have: \[ \boxed{\mathcal{N}(B) \subseteq \mathcal{N}(A B)} \] --- ### **Part (b): Show that \(\mathcal{R}(A B) \subseteq \mathcal{R}(A)\).** --- ### **Step 1: Understand the ranges** - \(\mathcal{R}(A B) = \{ y \in \mathbb{R}^m \mid y = A B x \text{ for some } x \in \mathbb{R}^n \}\) - \(\mathcal{R}(A) = \{ y \in \mathbb{R}^m \mid y = A z, z \in \mathbb{R}^r \}\) --- ### **Step 2: Take an arbitrary vector in \(\mathcal{R}(A B)\)** Let \(y \in \mathcal{R}(A B)\). Then: \[ \exists x \in \mathbb{R}^n \quad \text{such that} \quad y = A B x \] Define: \[ z := B x \] Since \(z \in \mathbb{R}^r\), then: \[ y = A z \] This shows that \(y\) is in the range of \(A\). --- ### **Step 3: Final conclusion** Because the above argument holds for any \(y \in \mathcal{R}(A B)\), we have: \[ \boxed{\mathcal{R}(A B) \subseteq \mathcal{R}(A)} \] --- ## **Summary of Final Answers** \[ \boxed{ \begin{aligned} &\text{51.} \quad \mathcal{N}(A) \cap \mathcal{N}(B) \subseteq \mathcal{N}(A + B) \\ &\text{52.} \quad \begin{cases} \text{(a)} \quad \mathcal{N}(B) \subseteq \mathcal{N}(A B) \\ \text{(b)} \quad \mathcal{R}(A B) \subseteq \mathcal{R}(A) \end{cases} \end{aligned} } \] These inclusions follow directly from the definitions and properties of null spaces and ranges of linear transformations/matrices.