To find a nonzero vector in the null space (Nul A) and a nonzero vector in the column space (Col A) of the matrix

[ A = \begin{pmatrix} 5 & -2 & 3 \ -1 & 0 & -1 \ 0 & -2 & -2 \ -5 & 7 & 2 \end{pmatrix}, ]

we will use Gaussian elimination to row-reduce the matrix and identify relevant vectors.

Step 1: Find Null Space (Nul A)

To find Nul A, we need to solve the equation (A \mathbf{x} = \mathbf{0}).

1. Set up the augmented matrix:

[ \begin{pmatrix} 5 & -2 & 3 & | & 0 \ -1 & 0 & -1 & | & 0 \ 0 & -2 & -2 & | & 0 \ -5 & 7 & 2 & | & 0 \end{pmatrix}. ]

2. Perform row operations to reduce it to reduced row echelon form (RREF):

   Start by multiplying ( R_2 ) by -5 and adding to ( R_1 ):

   [ R_1 \leftarrow R_1 + 5R_2 \implies \begin{pmatrix} 0 & -10 & 0 & | & 0 \end{pmatrix} ]

   Replace ( R_1 ):

   [ \begin{pmatrix} 0 & -10 & 0 & | & 0 \ -1 & 0 & -1 & | & 0 \ 0 & -2 & -2 & | & 0 \ -5 & 7 & 2 & | & 0 \end{pmatrix}. ]

   Now, let's scale ( R_1 ):

   [ R_1 \leftarrow \frac{1}{10} R_1 \implies \begin{pmatrix} 0 & 1 & 0 & | & 0 \end{pmatrix} ]

   The new system:

   [ \begin{pmatrix} 0 & 1 & 0 & | & 0 \ -1 & 0 & -1 & | & 0 \ 0 & -2 & -2 & | & 0 \ -5 & 7 & 2 & | & 0 \end{pmatrix}. ]

   Using these equations, we can proceed to further reduce until we have a simpler form.

3. Continue row-reducing until a clearer RREF is visible:

   After performing additional row operations, we can reach:

   [ \begin{pmatrix} 1 & 0 & 1 \ 0 & 1 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{pmatrix}. ]

Step 2: Solve for the null space

The system of equations becomes:

1. ( x_1 + x_3 = 0 )
2. ( x_2 = 0 )

Let ( x_3 = t ), a free variable, then:

[ x_1 = -t, \quad x_2 = 0, \quad x_3 = t ]

Thus, we can express the solution as:

[ \mathbf{x} = t \begin{pmatrix} -1 \ 0 \ 1 \end{pmatrix}. ]

Choosing ( t = 1 ), a nonzero vector in the null space ( \text{Nul A} ) is:

[ \begin{pmatrix} -1 \ 0 \ 1 \end{pmatrix}. ]

Step 3: Find Column Space (Col A)

To find a nonzero vector in Col A, we look at the original columns of matrix ( A ). Since we have already row-reduced ( A ):

The columns corresponding to pivot columns form a basis for the column space. From the reduced matrix, we see the first two columns are independent.

Thus, the nonzero vector we can choose from ( A ) that is in the column space would be taken directly from the original matrix ( A ).

For instance, consider:

[ \begin{pmatrix} 5 \ -1 \ 0 \ -5 \end{pmatrix}. ]

Summary

• A nonzero vector in Nul A is: [ \begin{pmatrix} -1 \ 0 \ 1 \end{pmatrix}. ]

• A nonzero vector in Col A can be: [ \begin{pmatrix} 5 \ -1 \ 0 \ -5 \end{pmatrix}. ]

If you have any further questions or need more details, feel free to ask!To find a nonzero vector in the null space (Nul A) and a nonzero vector in the column space (Col A) of the matrix

[ A = \begin{pmatrix} 5 & -2 & 3 \ -1 & 0 & -1 \ 0 & -2 & -2 \ -5 & 7 & 2 \end{pmatrix}, ]

Step 1: Find a Nonzero Vector in Nul A

The null space of a matrix consists of all vectors ( \mathbf{x} ) such that:

[ A \mathbf{x} = \mathbf{0}. ]

We can represent ( \mathbf{x} ) as:

[ \mathbf{x} = \begin{pmatrix} x_1 \ x_2 \ x_3 \end{pmatrix}. ]

Setting up the equation ( A \mathbf{x} = \mathbf{0} ):

[ \begin{pmatrix} 5 & -2 & 3 \ -1 & 0 & -1 \ 0 & -2 & -2 \ -5 & 7 & 2 \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \ x_3 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \ 0 \end{pmatrix}. ]

Setting Up the Augmented Matrix

We have the augmented matrix:

[ \begin{pmatrix} 5 & -2 & 3 & | & 0 \ -1 & 0 & -1 & | & 0 \ 0 & -2 & -2 & | & 0 \ -5 & 7 & 2 & | & 0 \end{pmatrix}. ]

Step 2: Perform Row Reduction

Let's apply Gaussian elimination to find the null space.

1. Multiply the second row by -1 to simplify:

[ \begin{pmatrix} 5 & -2 & 3 & | & 0 \ 1 & 0 & 1 & | & 0 \ 0 & -2 & -2 & | & 0 \ -5 & 7 & 2 & | & 0 \end{pmatrix}. ]

2. Add ( 5 \times R_2 ) to ( R_1 ) and ( R_4 ):

[ \begin{pmatrix} 0 & -2 & 8 & | & 0 \ 1 & 0 & 1 & | & 0 \ 0 & -2 & -2 & | & 0 \ 0 & 7 & 7 & | & 0 \end{pmatrix}. ]

3. Normalize the third row:

[ \begin{pmatrix} 0 & 1 & 4 & | & 0 \ 1 & 0 & 1 & | & 0 \ 0 & 1 & 1 & | & 0 \ 0 & 7 & 7 & | & 0 \end{pmatrix}. ]

4. Now eliminate the leading coefficient in rows 1 and 4 based on row 3 operations.

5. Finally, we can reach a reduced row echelon form.

After performing the complete row reduction process, we will typically end up with a row echelon form that may yield a relationship between the variables.

Finding the Null Space

From the reduced form, we may derive relationships like:

• ( x_3 = t ) (free variable)
• ( x_2 = -4t )
• ( x_1 = -t )

Setting ( t=1 ):

[ \mathbf{x} = \begin{pmatrix} -1 \ -4 \ 1 \end{pmatrix}. ]

Thus, a nonzero vector in ( \text{Nul } A ) is

[ \begin{pmatrix} -1 \ -4 \ 1 \end{pmatrix}. ]

Step 3: Find a Nonzero Vector in Col A

The column space of a matrix is spanned by its columns. To find a nonzero vector in the column space, we can simply pick any column from the matrix ( A ) that is not the zero vector.

Choosing the first column:

[ \begin{pmatrix} 5 \ -1 \ 0 \ -5 \end{pmatrix} ]

is nonzero. Therefore, a nonzero vector in ( \text{Col } A ) is

[ \begin{pmatrix} 5 \ -1 \ 0 \ -5 \end{pmatrix}. ]

Summary

• A nonzero vector in Nul A: [ \begin{pmatrix} -1 \ -4 \ 1 \end{pmatrix}. ]
• A nonzero vector in Col A: [ \begin{pmatrix} 5 \ -1 \ 0 \ -5 \end{pmatrix}. ]