SOLUTION: 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 representa

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.