Orthogonal Vectors in R³: Basis and Linear Combination

Answer

The set of vectors is orthogonal because the dot product of every distinct pair equals zero. Since the set consists of three non-zero orthogonal vectors in $\mathbb{R}^3$, they are linearly independent and thus form an orthogonal basis, allowing $\mathbf{y} = [3, 4, 5]^T$ to be decomposed using orthogonal projections.

Explanation

1. Verify orthogonality using the dot product Let $\mathbf{u_1} = \begin{bmatrix} 2 \\ -1 \\ 2 \end{bmatrix}$, $\mathbf{u_2} = \begin{bmatrix} 1 \\ -2 \\ -2 \end{bmatrix}$, and $\mathbf{u_3} = \begin{bmatrix} -2 \\ -2 \\ 1 \end{bmatrix}$. We compute the dot product for each pair: $\mathbf{u_1} \cdot \mathbf{u_2} = (2)(1) + (-1)(-2) + (2)(-2) = 2 + 2 - 4 = 0$ This confirms that the first and second vectors are perpendicular. $\mathbf{u_1} \cdot \mathbf{u_3} = (2)(-2) + (-1)(-2) + (2)(1) = -4 + 2 + 2 = 0$ The first and third vectors are also perpendicular. $\mathbf{u_2} \cdot \mathbf{u_3} = (1)(-2) + (-2)(-2) + (-2)(1) = -2 + 4 - 2 = 0$ The second and third vectors are perpendicular, completing the proof of an orthogonal set.

2. Establish the basis property By theorem, any orthogonal set of non-zero vectors is linearly independent. Since we have a set of 3 linearly independent vectors in a 3-dimensional space ($\mathbb{R}^3$), the Basis Theorem implies they must span $\mathbb{R}^3$ and thus form a basis.

3. Compute coefficients using orthogonal projection To express $\mathbf{y} = \begin{bmatrix} 3 \\ 4 \\ 5 \end{bmatrix}$ as $c_1\mathbf{u_1} + c_2\mathbf{u_2} + c_3\mathbf{u_3}$, we use the formula $c_i = \frac{\mathbf{y} \cdot \mathbf{u_i}}{\mathbf{u_i} \cdot \mathbf{u_i}}$:

   For $c_1$: $\mathbf{y} \cdot \mathbf{u_1} = (3)(2) + (4)(-1) + (5)(2) = 6 - 4 + 10 = 12$ The dot product of the target vector and the first basis vector. $\mathbf{u_1} \cdot \mathbf{u_1} = 2^2 + (-1)^2 + 2^2 = 4 + 1 + 4 = 9$ The squared magnitude of the first basis vector. $c_1 = \frac{12}{9} = \frac{4}{3}$ The scalar weight for the first vector in the linear combination.

   For $c_2$: $\mathbf{y} \cdot \mathbf{u_2} = (3)(1) + (4)(-2) + (5)(-2) = 3 - 8 - 10 = -15$ The dot product of the target vector and the second basis vector. $\mathbf{u_2} \cdot \mathbf{u_2} = 1^2 + (-2)^2 + (-2)^2 = 1 + 4 + 4 = 9$ The squared magnitude of the second basis vector. $c_2 = \frac{-15}{9} = -\frac{5}{3}$ The scalar weight for the second vector in the linear combination.

   For $c_3$: $\mathbf{y} \cdot \mathbf{u_3} = (3)(-2) + (4)(-2) + (5)(1) = -6 - 8 + 5 = -9$ The dot product of the target vector and the third basis vector. $\mathbf{u_3} \cdot \mathbf{u_3} = (-2)^2 + (-2)^2 + 1^2 = 4 + 4 + 1 = 9$ The squared magnitude of the third basis vector. $c_3 = \frac{-9}{9} = -1$ The scalar weight for the third vector in the linear combination.

Final Answer

The vector written as a linear combination is: $\boxed{\begin{bmatrix} 3 \\ 4 \\ 5 \end{bmatrix} = \frac{4}{3}\begin{bmatrix} 2 \\ -1 \\ 2 \end{bmatrix} - \frac{5}{3}\begin{bmatrix} 1 \\ -2 \\ -2 \end{bmatrix} - 1\begin{bmatrix} -2 \\ -2 \\ 1 \end{bmatrix}}$

Common Mistakes

• Normalization Confusion: Students often forget to divide by $\mathbf{u_i} \cdot \mathbf{u_i}$ (the squared norm) and only calculate $\mathbf{y} \cdot \mathbf{u_i}$. This only works if the basis is orthonormal.
• Basis Theorem Oversight: Failing to mention that the vectors are non-zero; the internal dot products being zero implies independence only if no vector in the set is the zero vector.

Related Topics: Gram-Schmidt Process, Orthonormal Bases, Fourier Coefficients in Linear Algebra.