Questions: Let u₁ = [-2; 0; 0], u₂ = [-4; 2; 0], u₃ = [1; 3; -3] be a basis for R³. Use the Gram-Schmidt process to find an orthogonal basis under the Euclidean inner product. Orthogonal basis: v₁ = [-2; 0; 0], v₂ = [0; a; 0], v₃ = [0; b; c] a = Ex: 5, b = Ex: 5, c = Ex: 5

Let u₁ = [-2; 0; 0], u₂ = [-4; 2; 0], u₃ = [1; 3; -3] be a basis for R³. Use the Gram-Schmidt process to find an orthogonal basis under the Euclidean inner product.

Orthogonal basis: v₁ = [-2; 0; 0], v₂ = [0; a; 0], v₃ = [0; b; c]
a = Ex: 5, b = Ex: 5, c = Ex: 5

Transcript text: Let $\left\{\mathbf{u}_{1}=\left[\begin{array}{c}-2 \\ 0 \\ 0\end{array}\right], \mathbf{u}_{2}=\left[\begin{array}{c}-4 \\ 2 \\ 0\end{array}\right], \mathbf{u}_{3}=\left[\begin{array}{c}1 \\ 3 \\ -3\end{array}\right]\right\}$ be a basis for $\mathbb{R}^{3}$. Use the Gram-Schmidt process to find an orthogonal basis under the Euclidean inner product. Orthogonal basis: $\left\{\mathbf{v}_{1}=\left[\begin{array}{c}-2 \\ 0 \\ 0\end{array}\right], \mathbf{v}_{2}=\left[\begin{array}{l}0 \\ a \\ 0\end{array}\right], \mathbf{v}_{3}=\left[\begin{array}{l}0 \\ b \\ c\end{array}\right]\right\}$ \[ a=\mathrm{Ex}: 5 \quad b=\mathrm{Ex:5} \quad c=\mathrm{Ex}: 5 \]

Solution

Solution Steps

Step 1: Define the Given Vectors

Let the vectors be defined as follows: \[ \mathbf{u}_{1} = \begin{bmatrix} -2 \\ 0 \\ 0 \end{bmatrix}, \quad \mathbf{u}_{2} = \begin{bmatrix} -4 \\ 2 \\ 0 \end{bmatrix}, \quad \mathbf{u}_{3} = \begin{bmatrix} 1 \\ 3 \\ -3 \end{bmatrix} \]

Step 2: Apply the Gram-Schmidt Process

We apply the Gram-Schmidt process to the set of vectors $\{\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{u}_{3}\}$ to obtain an orthogonal basis.

The first vector is taken as is: \[ \mathbf{v}_{1} = \mathbf{u}_{1} = \begin{bmatrix} -2 \\ 0 \\ 0 \end{bmatrix} \]
For the second vector, we compute: \[ \mathbf{v}_{2} = \mathbf{u}_{2} - \text{proj}_{\mathbf{v}_{1}} \mathbf{u}_{2} \] where \[ \text{proj}_{\mathbf{v}_{1}} \mathbf{u}_{2} = \frac{\mathbf{u}_{2} \cdot \mathbf{v}_{1}}{\mathbf{v}_{1} \cdot \mathbf{v}_{1}} \mathbf{v}_{1} \] After calculations, we find: \[ \mathbf{v}_{2} = \begin{bmatrix} 0 \\ 2 \\ 0 \end{bmatrix} \]
For the third vector, we compute: \[ \mathbf{v}_{3} = \mathbf{u}_{3} - \text{proj}_{\mathbf{v}_{1}} \mathbf{u}_{3} - \text{proj}_{\mathbf{v}_{2}} \mathbf{u}_{3} \] After performing the necessary projections, we find: \[ \mathbf{v}_{3} = \begin{bmatrix} 0 \\ 0 \\ -3 \end{bmatrix} \]

Step 3: Extract the Components of the Orthogonal Basis

From the orthogonal basis obtained, we identify the components: \[ \mathbf{v}_{1} = \begin{bmatrix} -2 \\ 0 \\ 0 \end{bmatrix}, \quad \mathbf{v}_{2} = \begin{bmatrix} 0 \\ 2 \\ 0 \end{bmatrix}, \quad \mathbf{v}_{3} = \begin{bmatrix} 0 \\ 0 \\ -3 \end{bmatrix} \] Thus, we have: \[ a = 2, \quad b = 0, \quad c = -3 \]