. Let
A =


2 4 3
9 6 3
−1 3 4
1 1 1

 

2 points) Is it possible to find a nonzero vector v 6= 0 such that Ax = v is consistent, and v is in the nullspace of AT
? If it is, provide an example. Otherwise,
explain why not.


Let's break down the problem step-by-step:

We are given matrix \( A \) and need to determine whether there exists a nonzero vector \( v \) such that:

1. The equation \( A \mathbf{x} = \mathbf{v} \) is consistent (meaning it has at least one solution for \( \mathbf{x} \)).
2. \( \mathbf{v} \) is in the nullspace of \( A^T \), meaning \( A^T \mathbf{v} = 0 \).

### Step 1: Understanding the nullspace of \( A^T \)
The nullspace of \( A^T \) consists of all vectors \( \mathbf{v} \) such that:
\[
A^T \mathbf{v} = 0
\]
This means that if \( \mathbf{v} \) is in the nullspace of \( A^T \), multiplying \( A^T \) by \( \mathbf{v} \) will yield the zero vector.

### Step 2: Check the consistency of \( A \mathbf{x} = \mathbf{v} \)
For \( A \mathbf{x} = \mathbf{v} \) to be consistent, the vector \( \mathbf{v} \) must lie within the column space of \( A \). That is, \( \mathbf{v} \) must be a linear combination of the columns of \( A \). If \( \mathbf{v} \) does not lie in the column space of \( A \), then the system will be inconsistent.

### Step 3: Compatibility of \( v \) in both conditions
The main issue is that for \( A \mathbf{x} = \mathbf{v} \) to have a solution, \( \mathbf{v} \) must be in the column space of \( A \), while for \( \mathbf{v} \) to be in the nullspace of \( A^T \), it must satisfy \( A^T \mathbf{v} = 0 \). Generally, these two spaces—the column space of \( A \) and the nullspace of \( A^T \)—are orthogonal complements of each other.

This implies that no nonzero vector \( \mathbf{v} \) can simultaneously satisfy both conditions, because the column space of \( A \) and the nullspace of \( A^T \) do not overlap except at the zero vector.

### Conclusion:
No nonzero vector \( \mathbf{v} \) can satisfy both conditions. Therefore, it is not possible to find a nonzero vector \( \mathbf{v} \neq 0 \) such that \( A \mathbf{x} = \mathbf{v} \) is consistent and \( \mathbf{v} \) is in the nullspace of \( A^T \).

Would you like me to dive deeper into the calculation of the nullspace or column space for this matrix?

---

### Related Questions:
1. How is the nullspace of a matrix computed?
2. Can you explain the relationship between the nullspace of \( A^T \) and the column space of \( A \)?
3. What are the steps to determine if a system of linear equations is consistent?
4. How can we find the rank of a matrix and what does it tell us about the solution space?
5. What is the significance of the orthogonal complement in linear algebra?

#### Tip:
When solving systems of linear equations, always check the dimensions of the column space and null space. They can provide quick insight into whether the system is consistent or has unique or infinite solutions.

Let A = [[2, 4, 3], [9, 6, 3], [-1, 3, 4], [1, 1, 1]]. Is it possible to find a nonzero vector v ≠ 0 such that Ax = v is consistent, and v is in the nullspace of A^T? If it is, provide an example. Otherwise, explain why not.

Explore the conditions under which a nonzero vector v can satisfy both Ax = v and v being in the nullspace of A^T. Understand the relationship between the nullspace and column space, using matrix A = [[2, 4, 3], [9, 6, 3], [-1, 3, 4], [1, 1, 1]]. Suitable for undergraduate linear algebra students.

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