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This section will show that the eigendirections of F, introduced in Section 3.3, are the same as PNDDTPN and also the eigenvalues of F are equal to the eigenvalues of

−DDT plus 4 for i= 1. . . d

N and 0 otherwise. Where the dimension of the null space is

dN. Let the eigendecomposition of PNDDTPN equal J M JT, where J is a matrix of the eigenvectors and M has the eigenvalues along the diagonal and 0’s off the diagonal.

Let the eigendecompostion of DDT = Γ

sΛsΓTs. This implies that the eigendecom- postion of −DDT = Γ

s(−Λs)ΓTs. We wish to add 4×Id to −DDT in order to have all eigenvalues be greater than 0. It was shown in Section A.1 thatFf ull = Γs(4Id−Λs)ΓTs. The matrix PN(−DDT)PN = PN[Γs(−Λs)ΓTs]PN = J(−M)JT, while the matrix

F =PNΓs(cId−Λs)ΓsTPN. We wish to show that F =J(Z +M)JT, where

Z =    4IdN 0 0 0   

which implies that F has the eigenvalues 4Id−M for i = 1. . . dN and 0 otherwise and the eigenvectorsJ. This would mean thatPNDDTPN andF have the same eigenvectors, and the directions have simplicity scores which are shifted by a constant except for those that are 0.

We start with the characterization of F asPNΓs(−Λs+ 4Id)ΓTsPN.

PNΓs(−Λs+ 4Id)ΓTsPN

=PNΓs(−Λs)ΓTsPN +PNΓs(4Id)ΓTsPN =J(−M)JT +PN(4Id)PN

The last step in the above equations is done by noticing thatJ(−M)JT =PNΓs(−Λs)ΓTsPN by definition. Also notice that Γs(4Id)ΓTs = 4ΓsΓsT = 4Id. Next 4Id is replaced by J4JT, so

PNΓs(−Λs+ 4Id)ΓTsPN =J(−M)JT +PNJ4JTPN

By definition ofPN the firstdN eigenvectors ofJ will be in the null space and the others will not. Therefore the projection of Ji for i= 1. . . dN ontoN is equal to Ji. While the projection ofJi for i=dN + 1. . . d ontoN is equal to 0. Therefore

J(−M)JT +PNJ4JTPN

=J(−M)JT + [J1, J2, . . . , JdN,0, . . . ,0]4[J1, J2, . . . , JdN,0, . . . ,0]

T

=J(−M)JT + [J1, J2, . . . , JdN,0, . . . ,0]Z[J1, J2, . . . , JdN,0, . . . ,0]

T

In the above the zero vectors can be replace by [JdN+1. . . Jd] since the are being multiplied

by 0.

J(−M)JT + [J1, J2, . . . , JdN,0, . . . ,0]Z[J1, J2, . . . , JdN,0, . . . ,0]

T

=J(−M)JT +J ZJT

=J(−M+Z)JT

This shows the F = J(−M +Z)JT, so therefore F has eigenvectors J and eigenvalues which are the diagonal of (−M +Z). The value 4 can be replaced with any constant greater than 4 and the above would still hold.

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