## Step-by-Step Solution We are asked to find an **orthogonal basis** for the column space of: \[ A = \begin{bmatrix} -2 & -1 \\ -2 & 2 \\ -6 & 0 \end{bmatrix} \] Let's use the **Gram-Schmidt process** to orthogonalize the columns. --- ### **Step 1: Name the Columns** Let: \[ \mathbf{a}_1 = \begin{bmatrix} -2 \\ -2 \\ -6 \end{bmatrix}, \quad \mathbf{a}_2 = \begin{bmatrix} -1 \\ 2 \\ 0 \end{bmatrix} \] --- ### **Step 2: First Orthogonal Vector** Let \(\mathbf{v}_1 = \mathbf{a}_1\): \[ \mathbf{v}_1 = \begin{bmatrix} -2 \\ -2 \\ -6 \end{bmatrix} \] --- ### **Step 3: Orthogonalize the Second Vector** Remove the projection of \(\mathbf{a}_2\) onto \(\mathbf{v}_1\): \[ \text{proj}_{\mathbf{v}_1} \mathbf{a}_2 = \frac{\mathbf{a}_2 \cdot \mathbf{v}_1}{\mathbf{v}_1 \cdot \mathbf{v}_1} \mathbf{v}_1 \] #### Compute the dot products: - \(\mathbf{a}_2 \cdot \mathbf{v}_1 = (-1)(-2) + (2)(-2) + (0)(-6) = 2 - 4 + 0 = -2\) - \(\mathbf{v}_1 \cdot \mathbf{v}_1 = (-2)^2 + (-2)^2 + (-6)^2 = 4 + 4 + 36 = 44\) #### Compute projection: \[ \text{proj}_{\mathbf{v}_1} \mathbf{a}_2 = \frac{-2}{44} \begin{bmatrix} -2 \\ -2 \\ -6 \end{bmatrix} = \frac{-1}{22} \begin{bmatrix} -2 \\ -2 \\ -6 \end{bmatrix} = \begin{bmatrix} 2/22 \\ 2/22 \\ 6/22 \end{bmatrix} = \begin{bmatrix} 1/11 \\ 1/11 \\ 3/11 \end{bmatrix} \] #### Subtract to get \(\mathbf{v}_2\): \[ \mathbf{v}_2 = \mathbf{a}_2 - \text{proj}_{\mathbf{v}_1} \mathbf{a}_2 = \begin{bmatrix} -1 \\ 2 \\ 0 \end{bmatrix} - \begin{bmatrix} 1/11 \\ 1/11 \\ 3/11 \end{bmatrix} = \begin{bmatrix} -1 - 1/11 \\ 2 - 1/11 \\ 0 - 3/11 \end{bmatrix} = \begin{bmatrix} -12/11 \\ 21/11 \\ -3/11 \end{bmatrix} \] --- ### **Step 4: Orthogonal Basis** The orthogonal basis for the column space is: \[ \boxed{ \left\{ \begin{bmatrix} -2 \\ -2 \\ -6 \end{bmatrix}, \begin{bmatrix} -12 \\ 21 \\ -3 \end{bmatrix} \right\} } \] (The second vector has been scaled by 11 for clarity, but any nonzero scalar multiple is also valid.) --- ### **Summary Table** | Step | Vector | |---------------------------------|--------------------------------------------| | First basis vector | \( \begin{bmatrix} -2 \\ -2 \\ -6 \end{bmatrix} \) | | Second (orthogonalized) vector | \( \begin{bmatrix} -12 \\ 21 \\ -3 \end{bmatrix} \) | --- ## **Final Answer** \[ \boxed{ \left\{ \begin{bmatrix} -2 \\ -2 \\ -6 \end{bmatrix}, \begin{bmatrix} -12 \\ 21 \\ -3 \end{bmatrix} \right\} } \] is an **orthogonal basis** for the column space of the given matrix.

## Step-by-Step Solution We need to find an **orthogonal basis** for the column space of: \[ A = \begin{bmatrix} -2 & -1 \\ -2 & 2 \\ -6 & 0 \end{bmatrix} \] To achieve this, we will use the **Gram-Schmidt process** to orthogonalize the columns. --- ### Step 1: Define the Columns Let: \[ \mathbf{a}_1 = \begin{bmatrix} -2 \\ -2 \\ -6 \end{bmatrix}, \quad \mathbf{a}_2 = \begin{bmatrix} -1 \\ 2 \\ 0 \end{bmatrix} \] --- ### Step 2: First Orthogonal Vector Set \(\mathbf{v}_1\) as the first column: \[ \mathbf{v}_1 = \mathbf{a}_1 = \begin{bmatrix} -2 \\ -2 \\ -6 \end{bmatrix} \] --- ### Step 3: Orthogonalize the Second Vector To create the second orthogonal vector, we remove the projection of \(\mathbf{a}_2\) onto \(\mathbf{v}_1\): \[ \text{proj}_{\mathbf{v}_1} \mathbf{a}_2 = \frac{\mathbf{a}_2 \cdot \mathbf{v}_1}{\mathbf{v}_1 \cdot \mathbf{v}_1} \mathbf{v}_1 \] #### Compute the Dot Products 1. \( \mathbf{a}_2 \cdot \mathbf{v}_1 \): \[ \mathbf{a}_2 \cdot \mathbf{v}_1 = (-1)(-2) + (2)(-2) + (0)(-6) = 2 - 4 + 0 = -2 \] 2. \( \mathbf{v}_1 \cdot \mathbf{v}_1 \): \[ \mathbf{v}_1 \cdot \mathbf{v}_1 = (-2)^2 + (-2)^2 + (-6)^2 = 4 + 4 + 36 = 44 \] #### Compute the Projection \[ \text{proj}_{\mathbf{v}_1} \mathbf{a}_2 = \frac{-2}{44} \begin{bmatrix} -2 \\ -2 \\ -6 \end{bmatrix} = \frac{-1}{22} \begin{bmatrix} -2 \\ -2 \\ -6 \end{bmatrix} = \begin{bmatrix} \frac{2}{22} \\ \frac{2}{22} \\ \frac{6}{22} \end{bmatrix} = \begin{bmatrix} \frac{1}{11} \\ \frac{1}{11} \\ \frac{3}{11} \end{bmatrix} \] #### Subtract to Find \(\mathbf{v}_2\) \[ \mathbf{v}_2 = \mathbf{a}_2 - \text{proj}_{\mathbf{v}_1} \mathbf{a}_2 = \begin{bmatrix} -1 \\ 2 \\ 0 \end{bmatrix} - \begin{bmatrix} \frac{1}{11} \\ \frac{1}{11} \\ \frac{3}{11} \end{bmatrix} = \begin{bmatrix} -1 - \frac{1}{11} \\ 2 - \frac{1}{11} \\ 0 - \frac{3}{11} \end{bmatrix} = \begin{bmatrix} -\frac{12}{11} \\ \frac{21}{11} \\ -\frac{3}{11} \end{bmatrix} \] --- ### Step 4: Orthogonal Basis The orthogonal basis for the column space is: \[ \boxed{ \left\{ \begin{bmatrix} -2 \\ -2 \\ -6 \end{bmatrix}, \begin{bmatrix} -12 \\ 21 \\ -3 \end{bmatrix} \right\} } \] (The second vector can be scaled by 11 for clarity, resulting in integer values.) --- ### Summary Table | Step | Vector | |--------------------------------|--------------------------------------------| | First basis vector | \( \begin{bmatrix} -2 \\ -2 \\ -6 \end{bmatrix} \) | | Second (orthogonalized) vector | \( \begin{bmatrix} -12 \\ 21 \\ -3 \end{bmatrix} \) | --- ## Final Answer The orthogonal basis for the column space of the matrix is: \[ \boxed{ \left\{ \begin{bmatrix} -2 \\ -2 \\ -6 \end{bmatrix}, \begin{bmatrix} -12 \\ 21 \\ -3 \end{bmatrix} \right\} } \]