Linear Algebra With Applications

Otto Bretscher

$Math Studyset 91影视 Explanations$ Math

5th Edition 路 442 pages

ISBN: 9780321796974

Chapter 8: Q21E (page 412)

Find a polar decomposition $A = Q S$ as discussed in Exercise 20 for $A = [\begin{matrix} 6 & 2 \\ - 7 & 6 \end{matrix}]$ . Draw a sketch showing

Short Answer

Expert verified

$A = [\begin{matrix} \frac{4}{5} & \frac{3}{5} \\ - \frac{3}{5} & \frac{4}{5} \end{matrix}] [\begin{matrix} 9 & - 2 \\ - 2 & 6 \end{matrix}]$

Step by step solution

To find a polar decomposition

Given that,

$A = [\begin{matrix} 6 & 2 \\ - 7 & 6 \end{matrix}]$

Now from Example 2 and Example 4 we have:

${鈭�}_{}^{} = [\begin{matrix} 10 & 0 \\ 0 & 5 \end{matrix}] V = [\begin{matrix} \frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}} \\ - \frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \end{matrix}] U = [\begin{matrix} \frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \\ - \frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}} \end{matrix}]$

To find a polar decomposition

Hence from the Exercise 20 we have:

$Q = U V^{t} = [\begin{matrix} \frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \\ - \frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}} \end{matrix}] [\begin{matrix} \frac{2}{\sqrt{5}} & - \frac{1}{\sqrt{5}} \\ \frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \end{matrix}] = [\begin{matrix} \frac{4}{5} & \frac{3}{5} \\ - \frac{3}{5} & \frac{4}{5} \end{matrix}] S = V {鈭�}_{}^{} V^{t} = [\begin{matrix} \frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}} \\ - \frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \end{matrix}] [\begin{matrix} 10 & 0 \\ 0 & 5 \end{matrix}] [\begin{matrix} \frac{2}{\sqrt{5}} & - \frac{1}{\sqrt{5}} \\ \frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \end{matrix}] = [\begin{matrix} 9 & - 2 \\ - 2 & 6 \end{matrix}]$

Thus the required polar decomposition of A is given by:

$A = [\begin{matrix} \frac{4}{5} & \frac{3}{5} \\ - \frac{3}{5} & \frac{4}{5} \end{matrix}] [\begin{matrix} 9 & - 2 \\ - 2 & 6 \end{matrix}]$

Final proof

$A = [\begin{matrix} \frac{4}{5} & \frac{3}{5} \\ - \frac{3}{5} & \frac{4}{5} \end{matrix}] [\begin{matrix} 9 & - 2 \\ - 2 & 6 \end{matrix}]$

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91影视

Find a polar decomposition $A = Q S$ as discussed in Exercise 20 for $A = [\begin{matrix} 6 & 2 \\ - 7 & 6 \end{matrix}]$ . Draw a sketch showing

Short Answer

Step by step solution

To find a polar decomposition

To find a polar decomposition

Final proof

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91影视

Find a polar decomposition A=QSas discussed in Exercise 20 forA=[62-76]. Draw a sketch showing

Short Answer

Step by step solution

To find a polar decomposition

To find a polar decomposition

Final proof

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Probability and Statistics

Statistics

Decision Maths

Applied Mathematics

Theoretical and Mathematical Physics

Pure Maths

Study anywhere. Anytime. Across all devices.

Find a polar decomposition $A = Q S$ as discussed in Exercise 20 for $A = [\begin{matrix} 6 & 2 \\ - 7 & 6 \end{matrix}]$ . Draw a sketch showing