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.