advertisement

Math 2270 Quiz 9 1. Find an orthonormal basis for the kernel of the matrix 1 1 1 1 A= . 1 −1 −1 1 We already know how to find a basis for the kernel of A. First we’ll do this, then we’ll use the Gramm-Schmidt process to turn that basis into an orthonormal basis. We row-reduce A as 1 1 1 1 1 1 1 1 1 0 0 1 ∼ ∼ . 1 −1 −1 1 0 −2 −2 0 0 1 1 0 This implies that the homogeneous system has the solution x1 = −x4 . x2 = −x3 Let x3 = s and x4 = t. Then the solution vector can be written as x1 −t 0 x2 −s −1 x3 = s = s 1 + t x4 t 0 Let −1 0 . 0 1 −1 0 0 −1 ~v1 = 1 and ~v2 = 0 . 1 0 √ √ Then {~v1 , ~v2 } is a basis for ker(A). We have that k~v1 k = 1 + 1 = 2, so we take 0 − √1 2 ~u1 = √1 . 2 0 To find ~u2 , we find first ~v2⊥ ⊥ span(~u1 ), 0 −1 − √1 0 − (0) 1 2 = ~v2⊥ = ~v2 − (~u1 · ~v2 )~u1 = √ 0 2 1 0 √ √ We see that k~v2⊥ k = 1 + 1 = 2, so √1 − 2 −1 0 1 1 ⊥ 0 = . ~u2 = ⊥ ~v2 = √ k~v2 k 2 0 0 √1 1 2 Thus an orthonormal basis for ker(A) is 0 − √1 12 √ 2 0 − √12 0 . , 0 √1 2 −1 0 . 0 1 2. Suppose that A and B are orthogonal n × n matrices. Is the matrix AB also orthogonal? Justify your answer. Since A and B are orthogonal, we have that A−1 = AT and B −1 = B T , so (AB)(AB)T = (AB)(B T AT ) = ABB T AT = AIn AT = AAT = In , which implies that (AB)−1 = (AB)T and that AB is orthogonal.