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Q59E

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Found in: Page 109

### Linear Algebra With Applications

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

# TRUE OR FALSE?If a matrix ${\mathbit{A}}{\mathbf{=}}\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ represents the orthogonal projection onto a line L , then the equation ${\mathbit{a}}^{\mathbf{2}}\mathbf{+}{\mathbit{b}}^{\mathbf{2}}\mathbf{+}{\mathbit{c}}^{\mathbf{2}}\mathbf{+}{\mathbit{d}}^{\mathbf{2}}\mathbf{=}{\mathbf{1}}$ must hold.

The given statement is false.

See the step by step solution

## Step 1: Consider the parameters

The orthogonal projection of matrix is of form $\left(\begin{array}{cc}{u}_{1}^{2}& {u}_{1}{u}_{2}\\ {u}_{1}{u}_{2}& {u}_{2}^{2}\end{array}\right)$ with $\stackrel{\to }{u}=\left[\begin{array}{c}{u}_{1}\\ {u}_{2}\end{array}\right]$. Thus, if

${\mathrm{a}}^{2}+{\mathrm{b}}^{2}+{\mathrm{c}}^{2}+{\mathrm{d}}^{2}=1\mathrm{then}$

${\begin{array}{cc}{u}_{1}^{4}+& \left({u}_{1}{u}_{2}\right)\end{array}}^{2}+{\left({u}_{1}{u}_{2}\right)}^{2}+{u}_{2}^{4}=1\phantom{\rule{0ex}{0ex}}⇒{\left({u}_{1}{u}_{2}\right)}^{2}=1$

Consider the matrix.

$A=\left(\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ \mathrm{c}& \mathrm{d}\end{array}\right)$

## Step 2: Find the values of the elements

If $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ then,

The elements will be,

$a={u}_{1}^{2},d={u}_{2}^{2},b=c={u}_{1}{u}_{2}$

Compute, ${a}^{2}+{b}^{2}+{c}^{2}+{d}^{2}$

${a}^{2}+{b}^{2}+{c}^{2}+{d}^{2}={\left({u}_{1}^{2}\right)}^{2}+{\left({u}_{1}{u}_{2}\right)}^{2}+{\left({u}_{1}{u}_{2}\right)}^{2}+{\left({u}_{2}^{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{a}^{2}+{b}^{2}+{c}^{2}+{d}^{2}=\left({u}_{1}^{2}+{u}_{2}^{2}\right)\left({u}_{1}^{2}+{u}_{2}^{2}\right)$

Hence, the statement is false.