find 3 vector basis for null space A = <0, 2, -2, -1>, <0, 4, -4, -2>

Given the vectors \(\mathbf{A}_1 = \langle 0, 2, -2, -1 \rangle\) and \(\mathbf{A}_2 = \langle 0, 4, -4, -2 \rangle\), we are tasked with finding the null space of these vectors. The null space consists of all vectors \(\mathbf{x} = \langle x_1, x_2, x_3, x_4 \rangle\) that satisfy the matrix equation \( A \mathbf{x} = 0 \).

We represent the matrix \( A \) as:
\[
A = \begin{pmatrix}
0 & 2 & -2 & -1 \\
0 & 4 & -4 & -2
\end{pmatrix}
\]
and we want to solve for \(\mathbf{x}\) in the equation:
\[
A \mathbf{x} = \mathbf{0}.
\]
This gives the system of linear equations:
1. \( 2x_2 - 2x_3 - x_4 = 0 \)
2. \( 4x_2 - 4x_3 - 2x_4 = 0 \)

### Step 1: Solve the system
From the first equation:
\[
2x_2 - 2x_3 - x_4 = 0 \quad \Rightarrow \quad 2(x_2 - x_3) = x_4.
\]
Thus, we have \( x_4 = 2(x_2 - x_3) \).

Now let's check the second equation:
\[
4x_2 - 4x_3 - 2x_4 = 0.
\]
Substitute \( x_4 = 2(x_2 - x_3) \) into this:
\[
4x_2 - 4x_3 - 2[2(x_2 - x_3)] = 0 \quad \Rightarrow \quad 4x_2 - 4x_3 - 4x_2 + 4x_3 = 0.
\]
This simplifies to \( 0 = 0 \), which is true but gives no new information.

### Step 2: Parametrize the solution
We now know that \( x_4 = 2(x_2 - x_3) \), so the general solution for the null space can be written in terms of free variables \( x_2 \) and \( x_3 \). Let:
\[
x_1 = t, \quad x_2 = s, \quad x_3 = r.
\]
Then:
\[
x_4 = 2(s - r).
\]

Thus, the general form of a vector in the null space is:
\[
\mathbf{x} = \langle 0, s, r, 2(s - r) \rangle.
\]

### Step 3: Identify the basis
We can express the solution as a linear combination of the free variables \( s \) and \( r \). For \( s = 1 \) and \( r = 0 \), we get the vector:
\[
\mathbf{x}_1 = \langle 0, 1, 0, 2 \rangle.
\]
For \( s = 0 \) and \( r = 1 \), we get the vector:
\[
\mathbf{x}_2 = \langle 0, 0, 1, -2 \rangle.
\]

Thus, a basis for the null space is:
\[
\left\{ \mathbf{x}_1 = \langle 0, 1, 0, 2 \rangle, \mathbf{x}_2 = \langle 0, 0, 1, -2 \rangle \right\}.
\]
Since the system only has two free variables, the dimension of the null space is 2, not 3. Therefore, there are only 2 basis vectors for the null space.

Would you like more details or clarifications?

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Here are 5 related questions to consider:
1. How do you determine the dimension of the null space for a matrix?
2. What is the relationship between the rank of a matrix and its null space?
3. How does the concept of a null space apply to solving linear systems?
4. What is the geometric interpretation of a null space in vector spaces?
5. How can you find the null space of a larger matrix, say 3x3 or 4x4?

**Tip:** Always check the number of free variables in a linear system to determine the dimension of the null space!

Find 3 vector basis for null space A = <0, 2, -2, -1>, <0, 4, -4, -2>

Learn how to find the basis for the null space of a matrix formed by the vectors <0, 2, -2, -1> and <0, 4, -4, -2>. This detailed step-by-step solution covers key linear algebra concepts such as null space, systems of linear equations, and vector spaces. Suitable for college-level linear algebra students.

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