Short Answer

Complete the proof of Theorem 5.1.4: Orthogonal projection is linear transformation.

The transformation $T : ℝ^{n} \to ℝ^{n}$ with respect to the basis $\{{\vec{u}}_{1}, {\vec{u}}_{2}, . . ., {\vec{u}}_{m}\}$ is linear.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step by Step Solution Step 1: Determine the linear property of T

Consider the transformation $T : ℝ^{n} \to ℝ^{n}$ defined as $T (\vec{x}) = {\vec{x}}^{| |} = \sum_{i = 1}^{} ({\vec{u}}_{i} \cdot \vec{x}) {\vec{u}}_{i}$ where, $\{{\vec{u}}_{1}, {\vec{u}}_{2}, . . ., {\vec{u}}_{m}\}$ .

If the vector $\vec{x}$ is perpendicular $\vec{y}$ then $\vec{x} \cdot \vec{y} = 0$ .

Step 2: Proof

Assume $\vec{x}, \vec{y} \in ℝ^{n}$ and $α, β \in ℝ$ , simplify $T (α \vec{x} + β \vec{y}) = {(α \vec{x} + β \vec{y})}^{| |}$ as follows.

$\begin{array}{rcl} T (α \vec{x} + β \vec{y}) & = & {(α \vec{x} + β \vec{y})}^{| |} \\ T (α \vec{x} + β \vec{y}) & = & \sum_{i = 1}^{} ({\vec{u}}_{i} \cdot (α \vec{x} + β \vec{y})) {\vec{u}}_{i} \\ = & \sum_{i = 1}^{} ({\vec{u}}_{i} \cdot α \vec{x} + {\vec{u}}_{i} \cdot β \vec{y}) {\vec{u}}_{i} \\ T (α \vec{x} + β \vec{y}) & = & \sum_{i = 1}^{} (({\vec{u}}_{i} \cdot α \vec{x}) {\vec{u}}_{i} + ({\vec{u}}_{i} \cdot β \vec{y}) {\vec{u}}_{i}) \end{array}$

Further, simplify the equation as follows.

$\begin{array}{rcl} T (α \vec{x} + β \vec{y}) & = & \sum_{i = 1}^{} (({\vec{u}}_{i} \cdot α \vec{x}) {\vec{u}}_{i} + ({\vec{u}}_{i} \cdot β \vec{y}) {\vec{u}}_{i}) \\ T (α \vec{x} + β \vec{y}) & = & \sum_{i = 1}^{} (({\vec{u}}_{i} \cdot α \vec{x}) {\vec{u}}_{i}) + \sum_{i = 1}^{} (({\vec{u}}_{i} \cdot β \vec{y}) {\vec{u}}_{i}) \\ T (α \vec{x} + β \vec{y}) & = & α \{\sum_{i = 1}^{} (({\vec{u}}_{i} \cdot \vec{x}) {\vec{u}}_{i})\} + β \{\sum_{i = 1}^{} (({\vec{u}}_{i} \cdot \vec{y}) {\vec{u}}_{i})\} \\ T (α \vec{x} + β \vec{y}) & = & α \{T (\vec{x})\} + β \{T (\vec{y})\} \end{array}$

By the definition of linear transformation, the transformation $T$ is linear.

Hence, the transformation $T : ℝ^{n} \to ℝ^{n}$ with respect to the basis $\{{\vec{u}}_{1}, {\vec{u}}_{2}, . . ., {\vec{u}}_{m}\}$ is linear.

Find Study Materials for

Create Study Materials

Select your language

Linear Algebra With Applications

Answers without the blur.

Short Answer

Complete the proof of Theorem 5.1.4: Orthogonal projection is linear transformation.

Step by Step Solution

Step by Step Solution Step 1: Determine the linear property of T

Step 2: Proof

Most popular questions for Math Textbooks

Want to see more solutions like these?

Recommended explanations on Math Textbooks

Select your language

Linear Algebra With Applications

Answers without the blur.

Short Answer

Complete the proof of Theorem 5.1.4: Orthogonal projection is linear transformation.

Step by Step Solution

Step by Step Solution Step 1: Determine the linear property of T

Step 2: Proof

Most popular questions for Math Textbooks

Want to see more solutions like these?

Recommended explanations on Math Textbooks

Decision Maths

Calculus

Probability and Statistics

Statistics

Mechanics Maths

Geometry

Pure Maths