To find the orthogonal decomposition of $v$ with respect to $W$, we express $v$ as the sum of two vectors:

$v = \text{proj}_W(v) + \text{perp}_W(v)$

Where:

• $\text{proj}_W(v)$ is the projection of $v$ onto $W$.
• $\text{perp}_W(v)$ is the component of $v$ perpendicular to $W$.

Given: $v = \begin{bmatrix} 3 \\ -3 \end{bmatrix}, \quad W = \text{span}\left\{ \begin{bmatrix} 1 \\ 5 \end{bmatrix} \right\}$

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Step 1: Compute the projection $\text{proj}_W(v)$

The formula for the projection of $v$ onto $W$ is: $\text{proj}_W(v) = \frac{v \cdot w}{w \cdot w} w$

Here, $w = \begin{bmatrix} 1 \\ 5 \end{bmatrix}$.

1. Compute $v \cdot w$ (dot product of $v$ and $w$): $v \cdot w = \begin{bmatrix} 3 \\ -3 \end{bmatrix} \cdot \begin{bmatrix} 1 \\ 5 \end{bmatrix} = 3(1) + (-3)(5) = 3 - 15 = -12$

2. Compute $w \cdot w$ (dot product of $w$ with itself): $w \cdot w = \begin{bmatrix} 1 \\ 5 \end{bmatrix} \cdot \begin{bmatrix} 1 \\ 5 \end{bmatrix} = 1^2 + 5^2 = 1 + 25 = 26$

3. Compute $\frac{v \cdot w}{w \cdot w}$: $\frac{v \cdot w}{w \cdot w} = \frac{-12}{26} = -\frac{6}{13}$

4. Multiply $w$ by $\frac{v \cdot w}{w \cdot w}$: $\text{proj}_W(v) = -\frac{6}{13} \begin{bmatrix} 1 \\ 5 \end{bmatrix} = \begin{bmatrix} -\frac{6}{13} \\ -\frac{30}{13} \end{bmatrix}$

Thus: $\text{proj}_W(v) = \begin{bmatrix} -\frac{6}{13} \\ -\frac{30}{13} \end{bmatrix}$

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Step 2: Compute $\text{perp}_W(v)$

The perpendicular component is: $\text{perp}_W(v) = v - \text{proj}_W(v)$

1. Subtract $\text{proj}_W(v)$ from $v$: $\text{perp}_W(v) = \begin{bmatrix} 3 \\ -3 \end{bmatrix} - \begin{bmatrix} -\frac{6}{13} \\ -\frac{30}{13} \end{bmatrix}$

2. Perform the subtraction: $\text{perp}_W(v) = \begin{bmatrix} 3 + \frac{6}{13} \\ -3 + \frac{30}{13} \end{bmatrix} = \begin{bmatrix} \frac{39}{13} + \frac{6}{13} \\ -\frac{39}{13} + \frac{30}{13} \end{bmatrix} = \begin{bmatrix} \frac{45}{13} \\ -\frac{9}{13} \end{bmatrix}$

Thus: $\text{perp}_W(v) = \begin{bmatrix} \frac{45}{13} \\ -\frac{9}{13} \end{bmatrix}$

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Final Answer:

$\text{proj}_W(v) = \begin{bmatrix} -\frac{6}{13} \\ -\frac{30}{13} \end{bmatrix}, \quad \text{perp}_W(v) = \begin{bmatrix} \frac{45}{13} \\ -\frac{9}{13} \end{bmatrix}$

Would you like more details or further clarification?

Related Questions:

1. How is the dot product calculated for two vectors?
2. What are the properties of projections in linear algebra?
3. How is the span of a set of vectors defined?
4. What is the geometric meaning of $\text{perp}_W(v)$?
5. Can you generalize the orthogonal decomposition for higher-dimensional spaces?

Tip:

Always verify your projection calculations by ensuring $\text{proj}_W(v)$ lies in the span of $W$.