-8 2. Let vi-20 6 0 be a basis for a 2D subspace of R and let V be the matrix 4
ID: 3138177 • Letter: #
Question
-8 2. Let vi-20 6 0 be a basis for a 2D subspace of R and let V be the matrix 4 whose columns are defined by vi and v2 b. (2 points) Compute A - vv and B-Vv (3 points) Let U be a matrix with orthonormal columns ui and u2 that span the same subspace of R as is spanned by the columns of V. Is U an orthogonal matrix? c. d. (3 points) Compute C UUT and D- U'U. How are these results similar to and different from the results A and B from 2b) above? e. (2 points) Compute aAy ande-Cy where yWhat do you notice about a and c?Explanation / Answer
2. a. Since v1.v2 = 3*(-8)+0*0+4*6 =0, hence V has orthogonal columns.
b. A = VVT =
73
0
-36
0
0
0
-36
0
52
B = VT V =
25
0
0
100
c. An orthogonal matrix is necessarily a square matrix. Therefore, U is not an orthogonal matrix.
d. U =
3/5
-4/5
0
0
4/5
3/5
so that C = UUT=
1
0
0
0
0
0
0
0
1
and D = UTU =
1
0
0
1
Both A and C are 3x3 matrices, but have different entries except that the 2nd row of both A and C is a zero row. Also, both B and D aare 2x2 matrices , but have different entries except that a12 and a21 entries in both the matrices are 0.
e. a = A(1,1,1)T = (37,0,16)T and c = C(1,1,1)T = (1,0,1)T.
f.
Let M = [u1,u2,c] =
3/5
-4/5
1
0
0
0
4/5
3/5
1
The RREF of M is
1
0
7/5
0
1
-1/5
0
0
0
Hence c = (7/5)u1-(1/5)u2.
g. proju1 y = [(y.u1)/(u1.u1)]u1 = [(3/5+0+4/5)/(9/25+0+16/25]u1 = (7/5)(3/5,0,4/5)T = (21/25,0,28/25)T and proju2 y = [(y.u2)/(u2.u2)]u2 = [(-4/5+0+3/5)/(16/25+0+9/25)]u2 = (-1/5)(-4/5,0,3/5)T = (4/25,0,-3/25)T . Then, proju1 y+ proju2 y = (21/25,0,28/25)T+(4/25,0,-3/25)T = (1,0,1)T. Also, y –(1,0,1)T = (0,1,0)T. Thus, y = (1,1,1)T = (1,0,1)T +(0,1,0)T. The first of these vectors is in span {u1,u2} and the second is orthogonal to it.
73
0
-36
0
0
0
-36
0
52
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