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Found in: Page 216

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

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

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

See the 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 $T\left(\stackrel{\to }{x}\right)={\stackrel{\to }{x}}^{||}=\sum _{i=1}^{}\left({\stackrel{\to }{u}}_{i}·\stackrel{\to }{x}\right){\stackrel{\to }{u}}_{i}$ where, $\left\{{\stackrel{\to }{u}}_{1},{\stackrel{\to }{u}}_{2},...,{\stackrel{\to }{u}}_{m}\right\}$.

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

## Step 2: Proof

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

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

Further, simplify the equation as follows.

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

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 $\left\{{\stackrel{\to }{u}}_{1},{\stackrel{\to }{u}}_{2},...,{\stackrel{\to }{u}}_{m}\right\}$ is linear.