find a basis for null space A = <0, 2, -2, -1>, <0, 4, -4, -2>

To find a basis for the null space of a matrix, we typically start by setting up the equation \( A \mathbf{x} = \mathbf{0} \), where \( A \) is the given matrix and \( \mathbf{x} \) is the vector of unknowns. However, in your case, it seems that the matrix has not been provided fully, but we can infer that \( A \) is made up of the two vectors given:

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

### Steps to find a basis for the null space:

1. Set up the system of equations represented by \( A \mathbf{x} = 0 \), where \( \mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} \).

\[
\begin{bmatrix}
0 & 2 & -2 & -1 \\
0 & 4 & -4 & -2
\end{bmatrix}
\begin{bmatrix}
x_1 \\ x_2 \\ x_3 \\ x_4
\end{bmatrix}
= \begin{bmatrix}
0 \\ 0
\end{bmatrix}
\]

This gives us two equations:

\[
2x_2 - 2x_3 - x_4 = 0
\]
\[
4x_2 - 4x_3 - 2x_4 = 0
\]

2. The second equation is simply twice the first, so we only need to solve the first equation:
\[
2x_2 - 2x_3 - x_4 = 0
\]
Simplifying:
\[
x_2 - x_3 - \frac{1}{2}x_4 = 0
\]
This can be rewritten as:
\[
x_2 = x_3 + \frac{1}{2}x_4
\]

3. Parametrize the solution:
- Let \( x_3 = t \) and \( x_4 = s \), where \( t \) and \( s \) are free variables.
- Then \( x_2 = t + \frac{1}{2}s \), and \( x_1 = 0 \) (since the first column of \( A \) is all zeros).

Thus, the solution vector \( \mathbf{x} \) can be written as:
\[
\mathbf{x} = \begin{bmatrix} 0 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 0 \\ t + \frac{1}{2}s \\ t \\ s \end{bmatrix}
\]
This can be split into two independent vectors by setting \( t = 1, s = 0 \) and \( t = 0, s = 1 \):
\[
\mathbf{x} = t \begin{bmatrix} 0 \\ 1 \\ 1 \\ 0 \end{bmatrix} + s \begin{bmatrix} 0 \\ \frac{1}{2} \\ 0 \\ 1 \end{bmatrix}
\]

4. Therefore, a basis for the null space is:
\[
\left\{ \begin{bmatrix} 0 \\ 1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ \frac{1}{2} \\ 0 \\ 1 \end{bmatrix} \right\}
\]

This is the set of vectors that forms a basis for the null space of \( A \).

Would you like a detailed explanation on any of these steps?

### Related Questions:
1. What is the dimension of the null space (nullity) for this matrix?
2. How do you find the row space and column space of a matrix?
3. Can a matrix have more than one basis for its null space?
4. What is the significance of the rank-nullity theorem in this context?
5. How does the number of free variables relate to the nullity of a matrix?

**Tip:** When dealing with free variables in null space, each free variable corresponds to a dimension in the null space, and the number of free variables gives the nullity.

Explore how to find a basis for the null space of a 2x4 matrix A = [0, 2, -2, -1], [0, 4, -4, -2] using linear algebra methods. Follow step-by-step solutions involving vector space concepts and the rank-nullity theorem, suitable for college-level math students.

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