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Extra info for An Introduction to Multicomplex Spaces and Functions
Conceptually this result is expected since the vectors a1 and a2 are perpendicular and when we project a given vector onto a1 we produce a vector that will still be perpendicular to a2 . Projecting this perpendicular vector onto a2 will result in a zero vector. Problem 6 From Problem 5 we have that P1 1 1 −2 P1 = 9 −2 and P2 given by given by −2 −2 4 4 4 4 4 4 −2 1 4 −2 P2 = 4 9 −2 −2 1 and finally P3 given by P3 = a3 aT3 aT3 a3 Then we have that = 4 1 −2 = 9 4 1 1 1 0 so P1 0 = 9 0 0 4 1 1 so P2 0 = 4 9 −2 0 2 1 −1 2 −1 2 4+1+4 2 −2 4 4 1 1 1 −2 so P3 0 = −2 .
Thus the (COL)((ROW )T decomposition would take the first r = 2 columns of E −1 with the first r = 2 rows of R . e. RB = [R R] , and the elimination matrix is the same. 4 (The Complete Solution to Ax = b) Problem 1 Let our augmented matrix A be, then with 1 3 3 1 A= 2 6 9 5 1 −3 3 5 E21 1 0 0 = −2 1 0 1 0 1 we have 1 3 3 1 E21 A = 0 0 3 3 0 0 6 6 continuing by dividing by the appropriate pivots and eliminating the elements below and above each pivot we have 1 3 0 −5 1 3 0 −5 1 3 3 1 E21 A = 0 0 3 3 ⇒ 0 0 0 0 ⇒ 0 0 1 0 0 0 6 6 0 0 1 1 0 0 0 0 From this expression we recognize the pivot variables of x1 and x3 .
When they are the same the norm (squared) of that column results. Problem 23 The lines 3x + y = b1 and 6x + 2y = b2 are parallel. They are the same line if 2b1 = b2 . Then [b1 , b2 ]T is perpendicular to the left nullspace of A= or 3 1 6 2 −2 . Note we can check that this vector is an element of the left nullspace by computing 1 −2 1 b1 b2 = −2b1 + b2 = −2b1 + 2b2 = 0 The nullspace of the matrix is the line 3x + y = 0. One vector in this nullspace is [−1, 3]T . Problem 24 Part (a): As discussed in the book if two subspaces are orthogonal then they can only meet at the origin.