Question

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

Short answer

The transformation $T:{\mathrm{ℝ}}^{\mathrm{n}}\to {\mathrm{ℝ}}^{\mathrm{n}}$ with respect to the basis$\left\{{\overrightarrow{u}}_1,{\overrightarrow{u}}_2,...,{\overrightarrow{u}}_m\right\}$ is linear.

Step-by-step solution

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

Consider the transformation $T:{\mathrm{ℝ}}^n\to {\mathrm{ℝ}}^n$ defined as $T\left(\overrightarrow{x}\right)={\overrightarrow{x}}^{\left|\right|}=\sum _{i=1}^{}\left({\overrightarrow{u}}_i·\overrightarrow{x}\right){\overrightarrow{u}}_i$ where, $\left\{{\overrightarrow{u}}_1,{\overrightarrow{u}}_2,...,{\overrightarrow{u}}_m\right\}$.

If the vector $\overrightarrow{x}$ is perpendicular $\overrightarrow{y}$ then $\overrightarrow{x}·\overrightarrow{y}=0$.

Proof

Assume $\overrightarrow{x},\overrightarrow{y}\in {\mathrm{ℝ}}^n$ and $\alpha ,\beta \in \mathrm{ℝ}$, simplify $T\left(\alpha \overrightarrow{x}+\beta \overrightarrow{y}\right)={\left(\alpha \overrightarrow{x}+\beta \overrightarrow{y}\right)}^{\left|\right|}$ as follows.

$\begin{array}{rcl}T\left(\alpha \overrightarrow{x}+\beta \overrightarrow{y}\right)& =& {\left(\alpha \overrightarrow{x}+\beta \overrightarrow{y}\right)}^{\left|\right|}\\ T\left(\alpha \overrightarrow{x}+\beta \overrightarrow{y}\right)& =& \sum _{i=1}^{}\left({\overrightarrow{u}}_i·\left(\alpha \overrightarrow{x}+\beta \overrightarrow{y}\right)\right){\overrightarrow{u}}_i\\ & =& \sum _{i=1}^{}\left({\overrightarrow{u}}_i·\alpha \overrightarrow{x}+{\overrightarrow{u}}_i·\beta \overrightarrow{y}\right){\overrightarrow{u}}_i\\ T\left(\alpha \overrightarrow{x}+\beta \overrightarrow{y}\right)& =& \sum _{i=1}^{}\left(\left({\overrightarrow{u}}_i·\alpha \overrightarrow{x}\right){\overrightarrow{u}}_i+\left({\overrightarrow{u}}_i·\beta \overrightarrow{y}\right){\overrightarrow{u}}_i\right)\end{array}$

Further, simplify the equation as follows.

$\begin{array}{rcl}T\left(\alpha \overrightarrow{x}+\beta \overrightarrow{y}\right)& =& \sum _{i=1}^{}\left(\left({\overrightarrow{u}}_i·\alpha \overrightarrow{x}\right){\overrightarrow{u}}_i+\left({\overrightarrow{u}}_i·\beta \overrightarrow{y}\right){\overrightarrow{u}}_i\right)\\ T\left(\alpha \overrightarrow{x}+\beta \overrightarrow{y}\right)& =& \sum _{i=1}^{}\left(\left({\overrightarrow{u}}_i·\alpha \overrightarrow{x}\right){\overrightarrow{u}}_i\right)+\sum _{i=1}^{}\left(\left({\overrightarrow{u}}_i·\beta \overrightarrow{y}\right){\overrightarrow{u}}_i\right)\\ T\left(\alpha \overrightarrow{x}+\beta \overrightarrow{y}\right)& =& \alpha \left\{\sum _{i=1}^{}\left(\left({\overrightarrow{u}}_i·\overrightarrow{x}\right){\overrightarrow{u}}_i\right)\right\}+\beta \left\{\sum _{i=1}^{}\left(\left({\overrightarrow{u}}_i·\overrightarrow{y}\right){\overrightarrow{u}}_i\right)\right\}\\ T\left(\alpha \overrightarrow{x}+\beta \overrightarrow{y}\right)& =& \alpha \left\{T\left(\overrightarrow{x}\right)\right\}+\beta \left\{T\left(\overrightarrow{y}\right)\right\}\\ & & \end{array}$

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

Hence, the transformation $T:{\mathrm{ℝ}}^n\to {\mathrm{ℝ}}^n$with respect to the basis $\left\{{\overrightarrow{u}}_1,{\overrightarrow{u}}_2,...,{\overrightarrow{u}}_m\right\}$is linear.