StudySmarter AI is coming soon!

- :00Days
- :00Hours
- :00Mins
- 00Seconds

A new era for learning is coming soonSign up for free

Suggested languages for you:

Americas

Europe

Q33E

Expert-verifiedFound in: Page 395

Book edition
5th

Author(s)
David C. Lay, Steven R. Lay and Judi J. McDonald

Pages
483 pages

ISBN
978-03219822384

**Construct a spectral decomposition of A from Example 2.**

The spectral decomposition of *A* is \(\left[ {\begin{aligned}{{}{}}6&{ - 2}&{ - 1}\\{ - 2}&6&{ - 1}\\{ - 1}&{ - 1}&5\end{aligned}} \right]\).

The eigenvalues of the matrix \(A = \left[ {\begin{aligned}{{}{}}6&{ - 2}&{ - 1}\\{ - 2}&6&{ - 1}\\{ - 1}&{ - 1}&5\end{aligned}} \right]\) are 8, 6, and 3. The matrix *P* is defined as:

\(\begin{aligned}{}P &= \left[ {\begin{aligned}{{}{}}{{{\bf{u}}_1}}&{{{\bf{u}}_2}}&{{{\bf{u}}_3}}\end{aligned}} \right]\\ &= \left[ {\begin{aligned}{{}{}}{ - \frac{1}{{\sqrt 2 }}}&{ - \frac{1}{{\sqrt 6 }}}&{\frac{1}{{\sqrt 3 }}}\\{\frac{1}{{\sqrt 2 }}}&{ - \frac{1}{{\sqrt 6 }}}&{\frac{1}{{\sqrt 3 }}}\\0&{\frac{2}{{\sqrt 6 }}}&{\frac{1}{{\sqrt 3 }}}\end{aligned}} \right]\end{aligned}\)

The spectral decomposition of *A* can be calculated as:

\(\begin{aligned}{}A &= {\lambda _1}{{\bf{u}}_1}{\bf{u}}_1^T + {\lambda _2}{{\bf{u}}_2}{\bf{u}}_2^T + {\lambda _3}{{\bf{u}}_3}{\bf{u}}_3^T\\ &= 8{{\bf{u}}_1}{\bf{u}}_1^T + 6{{\bf{u}}_2}{\bf{u}}_2^T + 6{{\bf{u}}_3}{\bf{u}}_3^T\\ &= 8\left[ {\begin{aligned}{{}{}}{ - \frac{1}{{\sqrt 2 }}}\\{\frac{1}{{\sqrt 2 }}}\\0\end{aligned}} \right]\left[ {\begin{aligned}{{}{}}{ - \frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 2 }}}&0\end{aligned}} \right] + 6\left[ {\begin{aligned}{{}{}}{ - \frac{1}{{\sqrt 6 }}}\\{ - \frac{1}{{\sqrt 6 }}}\\{\frac{2}{{\sqrt 6 }}}\end{aligned}} \right]\left[ {\begin{aligned}{{}{}}{ - \frac{1}{{\sqrt 6 }}}&{ - \frac{1}{{\sqrt 6 }}}&{\frac{2}{{\sqrt 6 }}}\end{aligned}} \right] + 3\left[ {\begin{aligned}{{}{}}{\frac{1}{{\sqrt 3 }}}\\{\frac{1}{{\sqrt 3 }}}\\{\frac{1}{{\sqrt 3 }}}\end{aligned}} \right]\left[ {\begin{aligned}{{}{}}{\frac{1}{{\sqrt 3 }}}&{\frac{1}{{\sqrt 3 }}}&{\frac{1}{{\sqrt 3 }}}\end{aligned}} \right]\end{aligned}\)

Solve further,

\(\begin{aligned}{}A &= 8\left[ {\begin{aligned}{{}{}}{\frac{1}{2}}&{ - \frac{1}{2}}&0\\{ - \frac{1}{2}}&{\frac{1}{2}}&0\\0&0&0\end{aligned}} \right] + 6\left[ {\begin{aligned}{{}{}}{\frac{1}{6}}&{\frac{1}{6}}&{ - \frac{2}{6}}\\{\frac{1}{6}}&{\frac{1}{6}}&{ - \frac{2}{6}}\\{ - \frac{2}{6}}&{ - \frac{2}{6}}&{\frac{4}{6}}\end{aligned}} \right] + 3\left[ {\begin{aligned}{{}{}}{\frac{1}{3}}&{\frac{1}{3}}&{\frac{1}{3}}\\{\frac{1}{3}}&{\frac{1}{3}}&{\frac{1}{3}}\\{\frac{1}{3}}&{\frac{1}{3}}&{\frac{1}{3}}\end{aligned}} \right]\\ &= \left[ {\begin{aligned}{{}{}}4&{ - 4}&0\\{ - 4}&4&0\\0&0&0\end{aligned}} \right] + \left[ {\begin{aligned}{{}{}}1&1&{ - 2}\\1&1&{ - 2}\\{ - 2}&{ - 2}&4\end{aligned}} \right] + \left[ {\begin{aligned}{{}{}}1&1&1\\1&1&1\\1&1&1\end{aligned}} \right]\\ &= \left[ {\begin{aligned}{{}{}}6&{ - 2}&{ - 1}\\{ - 2}&6&{ - 1}\\{ - 1}&{ - 1}&5\end{aligned}} \right]\end{aligned}\)

Thus, the spectral matrix of *A* is \(\left[ {\begin{aligned}{{}{}}6&{ - 2}&{ - 1}\\{ - 2}&6&{ - 1}\\{ - 1}&{ - 1}&5\end{aligned}} \right]\).

94% of StudySmarter users get better grades.

Sign up for free