Problem 5 Let A be an m x n matriz and X an n x 1 column vector. Let AX = 0 1. Explain why X is orthogonal to Row(A). That is: it is orthogonal to every vector in Row(A). (Hint: It is enough if it is orthogonal to all vectors in a basis of Row(A).) 2. Consider the following matriz, which is already in rref form. A = - 1 10-2 2 1 1 -3 4 Find one vector that is orthogonal to Row(A). (Just one vector is enough.) Solution 5:

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### Orthogonality and Linear Algebra: Problem 5

#### Problem Statement

Let \( A \) be an \( m \times n \) matrix and \( X \) an \( n \times 1 \) column vector. Consider the system:

\[ AX = 0 \]

1. **Explain why \( X \) is orthogonal to \( \text{Row}(A) \)**:  
   That is, it is orthogonal to every vector in \( \text{Row}(A) \).  
   *(Hint: It is enough if it is orthogonal to all vectors in a basis of \( \text{Row}(A) \).)*

2. **Consider the following matrix, which is already in reduced row echelon form (RREF):**

   \[
   A = \begin{bmatrix}
       1 & 0 & -2 & 2 \\
       0 & 1 & 1 & -3 \\
       0 & 0 & 1 & -4
   \end{bmatrix}
   \]

   Find one vector that is orthogonal to \( \text{Row}(A) \).  
   *(Just one vector is enough.)*

#### Solution

**Solution to Problem 5:**

---

**Additional Explanation:**

1. For the matrix \( A \) given:

\[ A = \begin{bmatrix}
    1 & 0 & -2 & 2 \\
    0 & 1 & 1 & -3 \\
    0 & 0 & 1 & -4
\end{bmatrix} \]

The problem asks us to find a vector \( X \) such that:

\[ AX = 0 \]

This will show that \( X \) is orthogonal to every row of \( A \), hence to \( \text{Row}(A) \), because any such \( X \) will be orthogonal to the span of the rows of \( A \).

---
Transcribed Image Text:### Orthogonality and Linear Algebra: Problem 5 #### Problem Statement Let \( A \) be an \( m \times n \) matrix and \( X \) an \( n \times 1 \) column vector. Consider the system: \[ AX = 0 \] 1. **Explain why \( X \) is orthogonal to \( \text{Row}(A) \)**: That is, it is orthogonal to every vector in \( \text{Row}(A) \). *(Hint: It is enough if it is orthogonal to all vectors in a basis of \( \text{Row}(A) \).)* 2. **Consider the following matrix, which is already in reduced row echelon form (RREF):** \[ A = \begin{bmatrix} 1 & 0 & -2 & 2 \\ 0 & 1 & 1 & -3 \\ 0 & 0 & 1 & -4 \end{bmatrix} \] Find one vector that is orthogonal to \( \text{Row}(A) \). *(Just one vector is enough.)* #### Solution **Solution to Problem 5:** --- **Additional Explanation:** 1. For the matrix \( A \) given: \[ A = \begin{bmatrix} 1 & 0 & -2 & 2 \\ 0 & 1 & 1 & -3 \\ 0 & 0 & 1 & -4 \end{bmatrix} \] The problem asks us to find a vector \( X \) such that: \[ AX = 0 \] This will show that \( X \) is orthogonal to every row of \( A \), hence to \( \text{Row}(A) \), because any such \( X \) will be orthogonal to the span of the rows of \( A \). ---
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