Question

Question: [M] A Landsat image with three spectral components was made of Homestead Air Force Base in Florida (after the base was hit by Hurricane Andrew in 1992). The covariance matrix of the data is shown below. Find the first principal component of the data, and compute the percentage of the total variance that is contained in this component.

\[S = \left[ {\begin{array}{*{20}{c}}{164.12}&{32.73}&{81.04}\\{32.73}&{539.44}&{249.13}\\{81.04}&{246.13}&{189.11}\end{array}} \right]\]

Short answer

The required percentage is \(75.8956\% \).

Step-by-step solution

Mean Deviation form and Covariance Matrix

The Mean Deviation form of any \(p \times N\)is given by:

\(B = \left[ {\begin{array}{*{20}{c}}{{{\hat X}_1}}&{{{\hat X}_2}}&{........}&{{{\hat X}_N}}\end{array}} \right]\)

Whose \(p \times p\)covariance matrix is:

\(S = \frac{1}{{N - 1}}B{B^T}\)

The Variance

From question, thecovariance matrix and themaximum eigenvalue we haveis:

\(\begin{array}{l}S = \left[ {\begin{array}{*{20}{c}}{164.12}&{32.73}&{81.04}\\{32.73}&{539.44}&{249.13}\\{81.04}&{246.13}&{189.11}\end{array}} \right]\\{\lambda _1} = 677.497\end{array}\)

The respective unit vector is:

\({u_1} = \left[ {\begin{array}{*{20}{c}}{0.129554}\\{0.874423}\\{0.467547}\end{array}} \right]\)

Now, the percentage of change in variance can be obtained as:

\[\begin{array}{c}\Delta = \frac{{{\lambda _1}}}{{tr\left( S \right)}} \times 100\\ = \frac{{677.4978}}{{164.12 + 539.44 + 189.11}} \times 100\\ = 75.8956\% \end{array}\]

Hence, this is the required answer.