Find an orthogonal basis for the column space of the matrix given on the right: $\begin{bmatrix} -1 & 5 & 6 \\ 3 & -6 & 2 \\ 2 & -1 & 6 \\ 1 & -5 & -2 \end{bmatrix}$

Neetesh Kumar | October 20, 2024

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This is the solution to Math2B Course: Linear Algebra
Assignment: Ch6 Section 4 Question Number 3
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Step-by-step solution:

1. Step 1: Let $\mathbf{v}_1$, $\mathbf{v}_2$, and $\mathbf{v}_3$ be the columns of the matrix. We apply the Gram-Schmidt process to these vectors to find an orthogonal basis.

Start with $\mathbf{u}_1 = \mathbf{v}_1$: $\mathbf{u}_1 = \begin{bmatrix} -1 \\ 3 \\ 2 \\ 1 \end{bmatrix}$

2. Step 2: Compute the projection of $\mathbf{v}_2$ onto $\mathbf{u}_1$: $\text{proj}_{\mathbf{u}_1} \mathbf{v}_2 = \frac{\mathbf{v}_2 \cdot \mathbf{u}_1}{\mathbf{u}_1 \cdot \mathbf{u}_1} \mathbf{u}_1$

First, compute the dot products: $\mathbf{v}_2 \cdot \mathbf{u}_1 = \begin{bmatrix} 5 \\ -6 \\ -1 \\ -5 \end{bmatrix} \cdot \begin{bmatrix} -1 \\ 3 \\ 2 \\ 1 \end{bmatrix} = 5(-1) + (-6)(3) + (-1)(2) + (-5)(1) = -5 - 18 - 2 - 5 = -30$

$\mathbf{u}_1 \cdot \mathbf{u}_1 = (-1)^2 + 3^2 + 2^2 + 1^2 = 1 + 9 + 4 + 1 = 15$

So, the projection is: $\text{proj}_{\mathbf{u}_1} \mathbf{v}_2 = \frac{-30}{15} \mathbf{u}_1 = -2 \begin{bmatrix} -1 \\ 3 \\ 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 2 \\ -6 \\ -4 \\ -2 \end{bmatrix}$

Now, subtract this projection from $\mathbf{v}_2$ to get $\mathbf{u}_2$: $\mathbf{u}_2 = \mathbf{v}_2 - \text{proj}_{\mathbf{u}_1} \mathbf{v}_2 = \begin{bmatrix} 5 \\ -6 \\ -1 \\ -5 \end{bmatrix} - \begin{bmatrix} 2 \\ -6 \\ -4 \\ -2 \end{bmatrix} = \begin{bmatrix} 3 \\ 0 \\ 3 \\ -3 \end{bmatrix}$

3. Step 3: Compute the projection of $\mathbf{v}_3$ onto $\mathbf{u}_1$ and $\mathbf{u}_2$.

First, the projection onto $\mathbf{u}_1$: $\text{proj}_{\mathbf{u}_1} \mathbf{v}_3 = \frac{\mathbf{v}_3 \cdot \mathbf{u}_1}{\mathbf{u}_1 \cdot \mathbf{u}_1} \mathbf{u}_1$

Dot products: $\mathbf{v}_3 \cdot \mathbf{u}_1 = \begin{bmatrix} 6 \\ 2 \\ 6 \\ -2 \end{bmatrix} \cdot \begin{bmatrix} -1 \\ 3 \\ 2 \\ 1 \end{bmatrix} = 6(-1) + 2(3) + 6(2) + (-2)(1) = -6 + 6 + 12 - 2 = 10$

Thus, $\text{proj}_{\mathbf{u}_1} \mathbf{v}_3 = \frac{10}{15} \mathbf{u}_1 = \frac{2}{3} \begin{bmatrix} -1 \\ 3 \\ 2 \\ 1 \end{bmatrix} = \begin{bmatrix} -\frac{2}{3} \\ 2 \\ \frac{4}{3} \\ \frac{2}{3} \end{bmatrix}$

Next, subtract this from $\mathbf{v}_3$: $\mathbf{v}_3 - \text{proj}_{\mathbf{u}_1} \mathbf{v}_3 = \begin{bmatrix} 6 \\ 2 \\ 6 \\ -2 \end{bmatrix} - \begin{bmatrix} -\frac{2}{3} \\ 2 \\ \frac{4}{3} \\ \frac{2}{3} \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix}$

Thus, the orthogonal basis for the column space is: $\mathbf{u}_1 = \begin{bmatrix} -\frac{1}{\sqrt{15}} \\ \frac{3}{\sqrt{15}} \\ \frac{2}{\sqrt{15}} \\ \frac{1}{\sqrt{15}} \end{bmatrix}, \quad \mathbf{u}_2 = \begin{bmatrix} \frac{1}{\sqrt{3}} \\ 0 \\ \frac{1}{\sqrt{3}} \\ -\frac{1}{\sqrt{3}} \end{bmatrix}, \quad \mathbf{u}_3 = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ 0 \\ 0 \\ \frac{1}{\sqrt{2}} \end{bmatrix}$

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