Solution 4.9 - Chemometrics: Data Analysis for the Laboratory and Chemical Plant

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  • Published: Jan 1, 2000
  • Channels: Chemometrics & Informatics

1. The scores and loadings are as follows.

2. Orthogonality in the scores and loadings is demonstrated by multiplying each PC with another and showing that the sum equals 0 (a very small residual is a consequence of computer rounding), for each method of scaling.

It is easy to demonstrate also that the sum of squares of the loadings for each PC and method of scaling equals 1.

3.The eigenvalues are as follows.

Raw

Centred

109.022

16.348

4.484

4.422

0.181

0.179

0.157

0.156

0.055

0.055

4. The sum of squares of the raw data equals 113.977, and that of the first five eigenvalues equals 113.897.

5. The means of the columns of the raw data, squared and multiplied by 10 are as follows.

6.291

5.606

9.986

17.509

23.886

18.320

9.201

1.963

The sum of the first five eigenvalues for the mean centred data is 21.159, and the sum of the eight values above is 92.760, leading to a total of 113.919, this is approximately the same as the sum of the first five eigenvalues of the raw data.

The reason is that the average of each column is subtracted from the raw data. The sum of the squares of the centred data equals the sum of squares of the raw data minus 10 (the number of objects) ´ sum of squares of each column. This can be shown algebraically as follows.

The equalities are exact when relating the sum of squares of the entire data, and approximate for five eigenvalues since they do not completely model the data.

6. It would look as if there are two significant components in the data. The centred data are easier to interpret because, for the uncentred data, the first PC dominates the analysis and is primarily a measure of magnitude in this case rather than variability. If one were looking at percentage variability rather than absolute variability, it would be possible to strongly underestimate the importance of the second eigenvalue using raw data.

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