Curve Resolution of two-factor systems

In their fundamental paper on curve resolution of two-component systems, Lawton and Sylvestre [7] studied a data matrix of spectra recorded during the elution of two constituents. One can decide either to estimate the pure spectra (and derive from them the concentration profiles) or the pure elution profiles (and derive from them the spectra) by factor analysis. Curve resolution, as developed by Lawton and Sylvestre, is based on the evaluation of the scores in the PC-space. Because the scores of the spectra in the PC-space defined by the wavelengths have a clearer structure (e.g. a line or a curve) than the scores of the elution profiles in the PC-space defined by the elution times, curve resolution usually estimates pure spectra. Thereafter, the pure elution profiles are estimated from the estimated pure spectra. Because no information on the specific order of the spectra is used, curve resolution is also applicable when the sequence of the spectra is not in a specific order. 

As explained before, the scores of the spectra can be plotted in the space defined by the two principal components of the data matrix. The appearance of the scores plot depends on the way the rows (spectra) and the columns have been normalized. If the spectra are not normalized, all spectra are situated in a plane (see Fig. 34.5). From the origin two straight lines depart, which are connected by a curved line. We have already explained that the straight line segments correspond with the pure spectra which are located in the wings of the elution bands (selective retention time 

Suppose that 15 spectra have been measured at 20 wavelengths and that after normalization on a sum equal to one they are compiled in a data matrix. Suppose also that by PCA the following score and loading matrices are obtained  

The scores and the loadings are plotted in Figs. 34.13 and 34.4. The fact that all absorbances of the pure spectra Sj and S2 should be non-negative means that the linear combination of the two PCs, s = TiV, + X2V2 should be non-negative for each element s, of s, given by  

Because the elements of v, all have equal sign, contrary to the elements of V2 which have mixed signs, x, /X2 values satisfying eq. (34.8) are found on the interval [/n,n], where m is equal to 

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