Solving Linear Quaternion Equations


by matwiz
Tags: equations, linear, quaternion, solving
matwiz
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Dec25-09, 01:10 AM
P: 5
Solving Linear Quaternion Equations

i,j,k, being "right-handed" unit quaternions
i',j',k' being the corresponding "left-handed" unit quaternions
q a general right quaternion variable q = w + xi + yj + zk,
c = c0 + ic1 + jc2 + kc3 constant right quaternion,

solve for q where,

3iq + 5kqj + (2i+k)qi = c ... EQ(1)




method,

move the known quaternions from R.H.S. to L.H.S
of variable q, by changing the "hand" of these
quaternions from right to left i.e. qj = j'q^
etc..

then,


3iq^ + 5kj'q^ + (2i+k)i'q^ = c^

( 3i + 5kj' + 2ii' + ki' )q^ = c^ ... EQ(2)

this has the form

hq^ = c^ ... EQ(3)

with

h = ( 3i + 5kj' + 2ii' + ki' )
h*R = ( -3i - 5kj' - 2ii' - ki' )
h*L = ( 3i - 5kj' - 2ii' - ki' )
h* = (h*R)*L = (h*L)*R = ( - 3i' + 5kj' + 2ii' + ki' )

the *R, *L, *, being Right-Hand Conjugate, Left-Hand Conjugate, Total Conjugate resp.

first multiply h by its total conjugate on the right to get,

hh* = (3i + 5kj' + 2ii' + ki').(3i + 5kj' + 2ii' + ki')*
= (3i + 5kj' + 2ii' + ki').(-3i + 5kj' + 2ii' + ki')
= +39 - 30jj' - 6ji' + 20jk'

now take the right conjugate of this,

(hh*)*R = ( +39 - 30jj' - 6ji' + 20jk' )*R
= ( +39 + 30jj' + 6ji' - 20jk' )

and multiply by h* from the left, so,

h*(hh*)*R = (-3i + 5kj' + 2ii' + ki').( +39 + 30jj' + 6ji' - 20jk' )
= ( + 39i - 12k - 22ii' - 130kj' + 21ki' + 195kj' + 20ij' )

Now multiply both sides of EQ(2) by this factor to get,


( + 39i - 12k - 22ii' - 130kj' + 21ki' + 195kj' + 20ij' )( 3i + 5kj' + 2ii' + ki' )q^ = ( + 39i - 12k - 22ii' - 130kj' + 21ki' + 195kj' + 20ij' )c^

i.e.

185q^ = ( + 39i - 12k - 22ii' - 130kj' + 21ki' + 195kj' + 20ij' )c^

Now that we have the real number "185" on the L.H.S of this equation, we can solve for q,

q^ = (1/185)( + 39i - 12k - 22ii' - 130kj' + 21ki' + 195kj' + 20ij' )c^

= (1/185)( + 39ic^ - 12kc^ - 22ii'c^ - 130kj'c^ + 21ki'c^ + 195kj'c^ + 20ij'c^ )

So, moving the left handed i',j',k', etc.. factors back over to the right of the constant c, we find,


q = ( + 39ic - 12kc - 22ici - 130kcj + 21kci + 195kcj + 20icj ) / 185


this can be verified by pluging the solution back into EQ(1).

Curious?

This general method is discussed in more detail on PDF page 143 of "Three Quaternion Papers" at archive.org

http://www.archive.org/details/Three...nions2006-2007
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