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Consider the vectors u1 = [1, 1, 1, 1], u2 = [0, 1, 1, 1], u3 = [0, 0, 1, 1] and u4 = [0, 0, 0, 1] in R4.
The set B = {u1, u2, u3, u4} is a basis for R4.
Find the unique representation of an arbitrary vector x = [x1, x2, x3, x4] in R4
as a linear combination of the vectors in B.
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Let e1 = (1,0,0,0),
e2 = (0,1,0,0),
e3 = (0,0,1,0),
e4 = (0,0,0,1)
be four standard basis vectors in R4.
Any arbitrary vector x = (x1,x2,x3,x4) of R4 is the linear combination of vectors e1, e2, e3, e4
x = x1*e1 + x2*e2 + x3*e3 + x4*e4.
Regarding the given vectors u1, u2, u3 and u4, we have, OBVIOUSLY,
u1 = e1 + e2 + e3 + e4, (1)
u2 = e2 + e3 + e4, (2)
u3 = e3 + e4, (3)
u4 = e4. (4)
Now, moving backward from equation (4) to equation(1), we can express vectors e1, e2, e3, e4 via vectors u1, u2, u3, u4:
e4 = u4 (5)
e3 = u3 - e4 = u3 - u4 (6)
e2 = u2 - e3 - e4 = u2 - (u3-u4) - u4 = u2 - u3 (7)
e1 = u1 - e2 - e3 - e4 = u1 - (u2 - u3) - (u3-u4) - u4 = u1 - u2 (8)
Next, any arbitrary vector x = x1*e1 + x2*e2 + x3*e3 +x4*e4 we can represent, substituting e-ith from (5) - (8)
x = x1*e1 + x2*e2 + x3*e3 +x4*e4 = x1*(u1-u2) + x2*(u2-u3) + x3*(u3-u4) + x4*u4 = regrouping =
= x1*u1 + (x2-x1)*u2 + (x3-x2)*u3 + (x4-x3)*u4. (9)
Thus formula (9) gives us that presentation of an arbitrary vector x via u1, u2, u3 and u4, which is requested by the problem.
So, the formula (9) provides the ANSWER.
Solved and explained in all details.