# Dimensionality Reduction using PCA, with SVD of correlation matrix

I have computed a correlation matrix from certain data set of dimension 6

The correlation matrix is:

 1.000000000000000 -0.142753907555000 -0.192138186332000 -0.523853268770000  0.124444699394000  0.002606132276000
-0.142753907555000  1.000000000000000  0.035741740609000  0.052631092693000  0.017746700663000 -0.361033166149000
-0.192138186332000  0.035741740609000  1.000000000000000  0.119582107782000 -0.346476373927000 -0.213738934536000
-0.523853268770000  0.052631092693000  0.119582107782000  1.000000000000000 -0.128989440426000  0.048428923563000
0.124444699394000  0.017746700663000 -0.346476373927000 -0.128989440426000  1.000000000000000 -0.434021957325000
0.002606132276000 -0.361033166149000 -0.213738934536000  0.048428923563000 -0.434021957325000  1.000000000000000


Then I applied SVD and got:

$U$

-0.568204264304213  0.191580010050053  0.344459905939842  0.057623813300714 -0.711866787458235  0.108229201374774
0.092203508694887 -0.529503080968478  0.114920527567713  0.783048200494938 -0.053598897055946  0.286148469345589
0.399921613786302 -0.136693414712789  0.669267980471650 -0.404735519465042  0.004693646229437  0.457848350968476
0.545763234721263 -0.103897122514854 -0.450027359344472 -0.130624377930601 -0.685583680847584  0.041659858939764
-0.440777193839364 -0.406362489876042 -0.425730028151024 -0.360866606588038  0.093306539446761  0.566046038136009
0.128697656758384  0.698794255931705 -0.190919707811538  0.269103577206925  0.107794763719302  0.612075746815743


$S$

1.757508326665900 0.000000000000000 0.000000000000000 0.000000000000000 0.000000000000000 0.000000000000000
0.000000000000000 1.561285064351230 0.000000000000000 0.000000000000000 0.000000000000000 0.000000000000000
0.000000000000000 0.000000000000000 1.108208836182120 0.000000000000000 0.000000000000000 0.000000000000000
0.000000000000000 0.000000000000000 0.000000000000000 0.829989535414321 0.000000000000000 0.000000000000000
0.000000000000000 0.000000000000000 0.000000000000000 0.000000000000000 0.469300557533083 0.000000000000000
0.000000000000000 0.000000000000000 0.000000000000000 0.000000000000000 0.000000000000000 0.273707679853350


$V^T$

-0.568204264304213  0.092203508694887  0.399921613786303  0.545763234721263 -0.440777193839364  0.128697656758384
0.191580010050053 -0.529503080968478 -0.136693414712789 -0.103897122514854 -0.406362489876042  0.698794255931705
0.344459905939843  0.114920527567713  0.669267980471650 -0.450027359344472 -0.425730028151024 -0.190919707811538
0.057623813300715  0.783048200494939 -0.404735519465042 -0.130624377930600 -0.360866606588038  0.269103577206925
-0.711866787458234 -0.053598897055946  0.004693646229437 -0.685583680847584  0.093306539446760  0.107794763719302
0.108229201374774  0.286148469345589  0.457848350968475  0.041659858939764  0.566046038136010  0.612075746815743


Now what do I do with this decomposition, Maybe i am wrong but $S$ has the eigenvalues, and $U$ has the eigenvectors. So how can I reduce dimension in this example? How would you interpret these results? Can you give a little guidance?

• Component scores = standardized data * U – ttnphns Nov 16 '12 at 19:21
• So I would take my original dataset and multiply it by $U$? and then sort the results, and the one with less value I can dismiss? – cMinor Nov 16 '12 at 19:27
• No sort needed because U's columns correspond to the eigenvalues which are sorted desdendingly. Just multiply Z*U and then keep first m columns of the result. Here you are with data reduced to m components. – ttnphns Nov 16 '12 at 19:34

• You said the first 4 because their $S$ values are $1.75$, $1.56$, $1.10$ and $0.82$ the others I can dismiss are $0.46$ and $0.27$ is that correct? the same goes for $U$? One more, so which attribute I am dropping $5$ and $6$? or how do you know that? – cMinor Nov 16 '12 at 19:42