Question

Construct a spectral decomposition of A from Example 3.

Short answer

The spectral decomposition of matrixA is \(7\left[ {\begin{aligned}{{}}{\frac{1}{2}}&0&{\frac{1}{2}}\\0&0&0\\{\frac{1}{2}}&0&{\frac{1}{2}}\end{aligned}} \right] + 7\left[ {\begin{aligned}{{}}{\frac{1}{{18}}}&{ - \frac{4}{{18}}}&{ - \frac{1}{{18}}}\\{ - \frac{4}{{18}}}&{\frac{{16}}{{18}}}&{\frac{4}{{18}}}\\{ - \frac{1}{{18}}}&{\frac{4}{{18}}}&{\frac{1}{{18}}}\end{aligned}} \right] - 2\left[ {\begin{aligned}{{}}{\frac{4}{9}}&{\frac{2}{9}}&{ - \frac{4}{9}}\\{\frac{2}{9}}&{\frac{1}{9}}&{ - \frac{2}{9}}\\{ - \frac{4}{9}}&{ - \frac{2}{9}}&{\frac{4}{9}}\end{aligned}} \right]\).

Step-by-step solution

Write given values from example 3

The eigenvalues of the matrix \(A = \left[ {\begin{aligned}{{}}3&{ - 2}&4\\{ - 2}&6&2\\4&2&3\end{aligned}} \right]\)are 7, 7, and \( - 2\). The matrix P is defined as:

\(\begin{aligned}{}P &= \left[ {\begin{aligned}{{}}{{{\bf{u}}_1}}&{{{\bf{u}}_2}}&{{{\bf{u}}_3}}\end{aligned}} \right]\\ &= \left[ {\begin{aligned}{{}}{\frac{1}{{\sqrt 2 }}}&{ - \frac{1}{{\sqrt {18} }}}&{ - \frac{2}{3}}\\0&{\frac{4}{{\sqrt {18} }}}&{ - \frac{1}{3}}\\{\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt {18} }}}&{\frac{2}{3}}\end{aligned}} \right]\end{aligned}\)

Find the spectral decomposition of A

The spectral decomposition of A can be calculated as:

\(\begin{aligned}{}A &= {\lambda _1}{{\bf{u}}_1}{\bf{u}}_1^T + {\lambda _2}{{\bf{u}}_2}{\bf{u}}_2^T + {\lambda _3}{{\bf{u}}_3}{\bf{u}}_3^T\\ &= 8{{\bf{u}}_1}{\bf{u}}_1^T + 6{{\bf{u}}_2}{\bf{u}}_2^T + 6{{\bf{u}}_3}{\bf{u}}_3^T\\ &= 7\left[ {\begin{aligned}{{}}{ - \frac{1}{{\sqrt 2 }}}\\0\\{\frac{1}{{\sqrt 2 }}}\end{aligned}} \right]\left[ {\begin{aligned}{{}}{\frac{1}{{\sqrt 2 }}}&0&{\frac{1}{{\sqrt 2 }}}\end{aligned}} \right] + 6\left[ {\begin{aligned}{{}}{ - \frac{1}{{\sqrt {18} }}}\\{\frac{4}{{\sqrt {18} }}}\\{\frac{1}{{\sqrt {18} }}}\end{aligned}} \right]\left[ {\begin{aligned}{{}}{ - \frac{1}{{\sqrt {18} }}}&{\frac{4}{{\sqrt {18} }}}&{\frac{1}{{\sqrt {18} }}}\end{aligned}} \right] - 2\left[ {\begin{aligned}{{}}{ - \frac{2}{3}}\\{ - \frac{1}{3}}\\{\frac{2}{3}}\end{aligned}} \right]\left[ {\begin{aligned}{{}}{ - \frac{2}{3}}&{ - \frac{1}{3}}&{\frac{2}{3}}\end{aligned}} \right]\\ &= 7\left[ {\begin{aligned}{{}}{\frac{1}{2}}&0&{\frac{1}{2}}\\0&0&0\\{\frac{1}{2}}&0&{\frac{1}{2}}\end{aligned}} \right] + 7\left[ {\begin{aligned}{{}}{\frac{1}{{18}}}&{ - \frac{4}{{18}}}&{ - \frac{1}{{18}}}\\{ - \frac{4}{{18}}}&{\frac{{16}}{{18}}}&{\frac{4}{{18}}}\\{ - \frac{1}{{18}}}&{\frac{4}{{18}}}&{\frac{1}{{18}}}\end{aligned}} \right] - 2\left[ {\begin{aligned}{{}}{\frac{4}{9}}&{\frac{2}{9}}&{ - \frac{4}{9}}\\{\frac{2}{9}}&{\frac{1}{9}}&{ - \frac{2}{9}}\\{ - \frac{4}{9}}&{ - \frac{2}{9}}&{\frac{4}{9}}\end{aligned}} \right]\end{aligned}\)

Thus, the spectral matrix of A is \(7\left[ {\begin{aligned}{{}}{\frac{1}{2}}&0&{\frac{1}{2}}\\0&0&0\\{\frac{1}{2}}&0&{\frac{1}{2}}\end{aligned}} \right] + 7\left[ {\begin{aligned}{{}}{\frac{1}{{18}}}&{ - \frac{4}{{18}}}&{ - \frac{1}{{18}}}\\{ - \frac{4}{{18}}}&{\frac{{16}}{{18}}}&{\frac{4}{{18}}}\\{ - \frac{1}{{18}}}&{\frac{4}{{18}}}&{\frac{1}{{18}}}\end{aligned}} \right] - 2\left[ {\begin{aligned}{{}}{\frac{4}{9}}&{\frac{2}{9}}&{ - \frac{4}{9}}\\{\frac{2}{9}}&{\frac{1}{9}}&{ - \frac{2}{9}}\\{ - \frac{4}{9}}&{ - \frac{2}{9}}&{\frac{4}{9}}\end{aligned}} \right]\).