Hi Yisheng,

It would probably be more polite if I were to quickly try to prove the
point to some degree, rather than just stating it. So...

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Given data vector Y and N linearly independent regressors R1, ... RN
(of the same length, less than or equal to N), there is a unique least
squares solution vector A to Y = RX where Y = RA + E (1), or:

Y = a1*R1 + a2*R2 + ... aN*RN + E

where E is minimal (as a sum of squared elements). This is presumably
known by us all (stating which almost assures that I mis-stated or
forgot some aspect, such is life).

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Now suppose that you want to modify regressor Ri by adding some
multiple m of Rj to it. Then the only effect to the solution will be
a corresponding change in the beta weight for regressor Rj.

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Proof:

First, rearrange the above equation into equivalent ones, by adding
and subtracting ai*m*Rj on the right side:

original:
Y = a1*R1 + ... + ai*Ri + ... + aj*Rj + ... + aN*RN + E

add and subtract ai*m*Rj:
Y = a1*R1 + ... + ai*Ri + (ai*m*Rj - ai*m*Rj) + ... + aj*Rj + ... + aN*RN + E

regroup:
Y = a1*R1 + ... + ai*(Ri+m*Rj) + ... + (aj-m*ai)*Rj + ... + aN*RN + E


So the ai term is the original ith beta times the modified regressor
Ri+m*Rj, and regressor Rj has a new beta weight (aj-m*ai). This is
equivalent to the original equation, and is still formed as a sum of
regressors in R. But if these newly grouped regressors are considered
to be a regression matrix S, then the new coefficients B must be a
least squares solution to Y = SX, where Y = SB + E (2).

Lastly we note that the solutions are unique (because the vectors in R
are linearly independent). If there were a different solution to (2)
then there would be a different solution to (1). So this is the unique
solution to (2).

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I guess it can be called a corollary that a similar argument could be
made for adding multiples of a more general subset of vectors in R to
some other subset of vectors in R, such as your adding multiples of the
constant and maybe linear polort regressors to the 6 motion regressors.

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More babbling than anticipated, but I have to wait for pictures to be
copied off my camera anyway... oops that's done. There are more
exciting things to do now. :)

- rick