imagine the values are vectors Given the set of vectors [2, -1, 2], [1, -2, -2], [-2, -2, 1] in R³, 1. Show that the vectors form an orthogonal set. 2. Without doing any further…

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.