Question

In each case, write \(\mathbf{x}\) as a linear combination of the orthogonal basis of the subspace \(U\). $$ \begin{array}{l} \text { a. } \mathbf{x}=(13,-20,15) ; U=\operatorname{span}\\{(1,-2,3),(-1,1,1)\\} \\ \text { b. } \mathbf{x}=(14,1,-8,5) \end{array} $$ \(U=\operatorname{span}\\{(2,-1,0,3),(2,1,-2,-1)\\}\)

Short answer

(a) \(\mathbf{x} = 7(1, -2, 3) - 6(-1, 1, 1)\); (b) \(\mathbf{x} = 3(2, -1, 0, 3) + 4(2, 1, -2, -1)\).

Step-by-step solution

Confirm Basis Orthogonality (Part a)

For part (a), we first check if the given basis vectors \((1, -2, 3)\) and \((-1, 1, 1)\) are orthogonal. Two vectors \(\mathbf{v_1}\) and \(\mathbf{v_2}\) are orthogonal if \(\mathbf{v_1} \cdot \mathbf{v_2} = 0\). Compute the dot product: \((1)(-1) + (-2)(1) + (3)(1) = -1 - 2 + 3 = 0\). Since the dot product is zero, the vectors are orthogonal.

Write \(\mathbf{x}\) as a Linear Combination (Part a)

Write \(\mathbf{x} = a(1, -2, 3) + b(-1, 1, 1)\). To find \(a\) and \(b\), solve the system of equations obtained by setting the linear combination equal to \(\mathbf{x} = (13, -20, 15)\):\[a(1) - b(1) = 13,a(-2) + b(1) = -20,a(3) + b(1) = 15.\] This simplifies to:1. \(a - b = 13\),2. \(-2a + b = -20\),3. \(3a + b = 15\).

Solve the Equations (Part a)

First, from equation (1) \(a = b + 13\). Substitute into equations (2) and (3):For equation (2):\[-2(b+13) + b = -20 \Rightarrow -2b - 26 + b = -20 \Rightarrow -b = 6 \Rightarrow b = -6.\]Now, for \(a\):\(a = -6 + 13 = 7\). Thus, \(\mathbf{x} = 7(1, -2, 3) - 6(-1, 1, 1)\).

Confirm Basis Orthogonality (Part b)

Check if the vectors \((2, -1, 0, 3)\) and \((2, 1, -2, -1)\) are orthogonal. Compute the dot product: \((2)(2) + (-1)(1) + (0)(-2) + (3)(-1) = 4 - 1 + 0 - 3 = 0\). Since the dot product is zero, the vectors are orthogonal.

Write \(\mathbf{x}\) as a Linear Combination (Part b)

Write \(\mathbf{x} = c(2, -1, 0, 3) + d(2, 1, -2, -1)\). To find \(c\) and \(d\), solve the system of equations obtained by setting the linear combination equal to \(\mathbf{x} = (14, 1, -8, 5)\):\[2c + 2d = 14,-c + d = 1,-2d = -8,3c - d = 5.\]

Solve the Equations (Part b)

From equation (3), \(-2d = -8 \Rightarrow d = 4\). Substitute \(d = 4\) into other equations:For equation (2):\(-c + 4 = 1 \Rightarrow -c = -3 \Rightarrow c = 3.\)Therefore, \(\mathbf{x} = 3(2, -1, 0, 3) + 4(2, 1, -2, -1)\).

Key concepts

Orthogonal Basis

An orthogonal basis is a set of vectors that are perpendicular to each other. This characteristic makes calculations, such as finding coefficients for linear combinations, easier. If two vectors within the basis are orthogonal, their dot product equals zero. For instance, in the original exercise, vectors such as

• (1, -2, 3) and (-1, 1, 1)
• (2, -1, 0, 3) and (2, 1, -2, -1)

are both orthogonal. To verify this, calculate their dot product and confirm it is zero. An orthogonal basis simplifies finding the representation of any vector in the subspace as a linear combination of the basis vectors, creating a straightforward path to solutions.

Dot Product

The dot product is a way to multiply two vectors, resulting in a scalar. This scalar indicates how much one vector is in the direction of another. It's also a crucial tool for checking orthogonality. Given vectors \( \mathbf{v_1} = (x_1, y_1, z_1) \) and \( \mathbf{v_2} = (x_2, y_2, z_2) \), their dot product is calculated as:\[\mathbf{v_1} \cdot \mathbf{v_2} = x_1x_2 + y_1y_2 + z_1z_2\]When two vectors are orthogonal, their dot product is zero. This is because their directions are completely independent, or perpendicular, in space. This concept was applied to determine the orthogonality in the original problem.

System of Equations

Systems of equations arise when solving for coefficients in linear combinations of vectors. In the exercise, once the orthogonality was confirmed, a system of equations allowed for finding the coefficients \( a, b, c, \) and \( d \). For example, to express a vector \( \mathbf{x} \) as a linear combination of orthogonal vectors, we equate \( \mathbf{x} = a\mathbf{v_1} + b\mathbf{v_2} \) and find \( a \) and \( b \) by solving equations like:\[a - b = 13\]\[-2a + b = -20 \]These equations are derived from matching each component of \( \mathbf{x} \) with its corresponding linear combination. Solving such systems involves algebraic manipulation to isolate variables, substituting back as needed, and thus provides the unique coefficients that reconstruct \( \mathbf{x} \) from the given basis.