Consider a system of equations Ax = b where A is of order m  m.
Let Ri 2 Rm; i = 1;   m be such that RT
i is the ith row of A. It is
given that each Ri is orthogonal to x for i = 1;   m with respect to the
standard inner product.
(i) Find b. (1 mark)
(ii) If S = fx 2 Rmj Ri is orthogonal to x for all i = 1;   mg where
orthogonality is with respect to the standard inner product. Then
prove that S is a subspace of Rm. (2 marks)
(iii) If rank (A) = m, nd dimension of S and S.

We are given a system of equations \( Ax = b \), where \( A \) is an \( m \times m \) matrix. Let \( R_i \in \mathbb{R}^m \) represent the rows of \( A \), and it is stated that each row \( R_i \) is orthogonal to the vector \( x \) with respect to the standard inner product. We are asked to solve for three parts of the problem:

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### (i) Find \( b \).

From the given information, we know that:
- \( R_i \cdot x = 0 \) for all \( i = 1, 2, \dots, m \), where \( R_i \) is the \( i \)-th row of \( A \), and \( \cdot \) denotes the standard dot product.
  
In matrix form, this means:
\[
Ax = \begin{bmatrix} R_1 \\ R_2 \\ \vdots \\ R_m \end{bmatrix} x = \begin{bmatrix} R_1 \cdot x \\ R_2 \cdot x \\ \vdots \\ R_m \cdot x \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{bmatrix} = 0
\]

Thus, \( b = 0 \).

#### Answer: \( b = 0 \).

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### (ii) Prove that \( S = \{ x \in \mathbb{R}^m \mid R_i \cdot x = 0 \text{ for all } i = 1, \dots, m \} \) is a subspace of \( \mathbb{R}^m \).

#### Subspace Proof:

To show that \( S \) is a subspace of \( \mathbb{R}^m \), we need to check the three subspace conditions:
1. **Zero vector belongs to \( S \)**:
   - For any \( R_i \), we have \( R_i \cdot 0 = 0 \), which satisfies the orthogonality condition. Thus, \( 0 \in S \).

2. **Closed under addition**:
   - Let \( x_1, x_2 \in S \). Then for all \( i \), \( R_i \cdot x_1 = 0 \) and \( R_i \cdot x_2 = 0 \). By the linearity of the dot product:
   \[
   R_i \cdot (x_1 + x_2) = R_i \cdot x_1 + R_i \cdot x_2 = 0 + 0 = 0
   \]
   Thus, \( x_1 + x_2 \in S \).

3. **Closed under scalar multiplication**:
   - Let \( x \in S \) and \( \lambda \in \mathbb{R} \). Then for all \( i \), \( R_i \cdot x = 0 \). Again, using the linearity of the dot product:
   \[
   R_i \cdot (\lambda x) = \lambda (R_i \cdot x) = \lambda \cdot 0 = 0
   \]
   Thus, \( \lambda x \in S \).

Since all three conditions hold, \( S \) is a subspace of \( \mathbb{R}^m \).

#### Answer: \( S \) is a subspace of \( \mathbb{R}^m \).

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### (iii) Find the dimension of \( S \) and \( S^\perp \).

Let’s proceed step by step:

- The set \( S \) consists of all vectors orthogonal to the rows of \( A \). Since \( R_i \cdot x = 0 \) for all \( i \), \( S \) is the null space (kernel) of \( A \).
  
- **If \( \text{rank}(A) = m \)**:  
  By the **Rank-Nullity Theorem**, we have:
  \[
  \text{rank}(A) + \text{nullity}(A) = m
  \]
  Since \( \text{rank}(A) = m \), the nullity of \( A \) is \( 0 \). Thus, \( \dim(S) = 0 \).

- **Dimension of \( S^\perp \)**:
  The space \( S^\perp \) is the orthogonal complement of \( S \). Since \( S \) is the null space (which has dimension 0), the orthogonal complement must span the entire space \( \mathbb{R}^m \). Therefore, \( \dim(S^\perp) = m \).

#### Answer:
- \( \dim(S) = 0 \)
- \( \dim(S^\perp) = m \)

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Let me know if you'd like more details on any of the parts or if you have any questions! Here are some follow-up questions to deepen your understanding:

1. Can you explain why \( S \) must be the null space of \( A \)?
2. How does the Rank-Nullity Theorem apply to subspaces?
3. What does it mean geometrically for a vector to be orthogonal to a set of vectors?
4. How do you interpret the rank of a matrix in terms of linear independence?
5. Why is the zero vector always part of any subspace?

Tip: Always check if a given set satisfies the subspace properties when proving subspaces.

Consider a system of equations Ax = b where A is of order m × m. Let Ri ∈ Rm; i = 1, 2, ... , m be such that RT_i is the ith row of A. It is given that each Ri is orthogonal to x for i = 1, 2, ... , m with respect to the standard inner product. (i) Find b. (ii) If S = {x ∈ Rm | Ri is orthogonal to x for all i = 1, ... , m} where orthogonality is with respect to the standard inner product, then prove that S is a subspace of Rm. (iii) If rank (A) = m, find the dimension of S and S^⊥.

Explore a detailed solution to a linear algebra problem involving a system of equations, orthogonal vectors, and subspace proof. Learn how to apply the Rank-Nullity Theorem and understand the concepts of orthogonality and subspaces. Suitable for undergraduate students.

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