Prof. E. Berlekamp Sep 21, 1999 Math 110 QUIZ SOLUTIONS QUESTION: 1) Without doing any permutations on the following real matrix, factor it into LU, where L is lower-triangular, and U is upper triangular. L should have all ones on its main diagonal. 2 1 0 0 0 3 2 1 0 0 0 3 2 1 0 0 0 3 2 1 0 0 0 3 2 ANSWER: 1 0 0 0 0 2 1 0 0 0 3/2 1 0 0 0 0 1/2 1 0 0 L = 0 6 1 0 0 U = 0 0 -4 1 0 0 0 -3/4 1 0 0 0 0 11/4 1 0 0 0 12/11 1 0 0 0 0 10/11 QUESTION: 2) The "Hamming" code is a binary vector space, defined as the nullspace of the following 4x8 binary matrix: 0 0 0 0 1 1 1 1 H = 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 1 0 0 1 0 1 1 0 The vectors in this code are called codewords. Any codeword is a possible message. Transmission noise might cause some bits of a message to be "erased", and replaced with "?"s. 2a) Suppose the received word is x0 0 x1 ? x2 1 x3 = 1 x4 ? x5 0 x6 ? x7 1 If possible, using any method you like, find the values of the bits in the three erased locations: x1, x4, and x6. SOLUTION 2a): 0 x1 0 0 0 0 1 1 1 1 1 0 0 0 1 1 0 0 1 1 1 = 0 0 1 0 1 0 1 0 1 x4 0 1 0 0 1 1 0 1 0 0 0 x6 1 whence 0 0 1 1 x1 0 0 0 1 1 1 1 0 0 1 x4 = 0 1 1 0 1 1 = 1 1 0 0 x6 0 0 1 1 1 0 0 0 1 1 1 0 1 0 0 1 1 Each of the 4 rows of the left matrix represents an equation in the 3 binary unknowns. The second equation says x6 = 1; the third equation says x1 = 0. The first and fourth equations say that x4 + x6 = 1, and since x6 = 1, x4 = 0. So, x1 0 x4 = 0 x6 1 QUESTION: 2b) If possible, find a set of 4 locations which, if erased, can be uniquely decoded. SOLUTION 2b) The sum of all 4 rows of the original H matrix is the all-ones row. Since this is in the row space of H, it means that every code vector must have an even number of bits of value 1. So no sum of 3 columns of H can be zero. Niether can any sum of 2 columns of H (because all columns are different), nor any single column of H (because no column is all-zero). The coordinates have conveniently been indexed according to the binary expansions of the first three rows of the H matrix. So if four positions have indices whose binary representations do not vector-sum to zero, then the corresponding four columns of H are linearly independent, and an erasure pattern which strikes those four coordinates can be uniquely decoded. For example, x0, x1, x2, and x4. Many other examples also work. QUESTION: 2c) If possible, find a set of 4 locations which, if erased, can NOT be uniquely decoded. SOLUTION 2c) x0, x1, x2, and x3 is such a set. The sum of the corresponding columns of H is zero. So no solution to this erasure pattern can be unique. Given any solution, changing all four erased bits would provide another solution. QUESTION: 3) Each part of this problem defines a vector space or subspace. Determine its dimension, and exhibit a basis. /* REFERENCES REFER TO THE TEXTBOOK */ 3a) /* page 88, # 2.3.13a */ The space of all vectors in R^4 whose components add to zero SOLUTION 3a) The answer is the null space of the 1x4 matrix 1111. Since the matrix has rank 1, its null space has dimension 3. Here is one basis: 1 0 0 v1 = 0 v2 = 1 v3 = 0 0 0 1 -1 -1 -1 (Of course, there are many other bases for this, and every other part of question 3.) Dimension = 3 QUESTION: 3b) /* page 129, # 2.11 */ The nullspace of the following real matrix: 1 2 0 1 2 1 U = 0 0 2 2 0 0 0 0 0 0 0 1 0 0 0 0 0 0 SOLUTION 3b) -2 -1 0 1 0 -1 N = 0 -1 0 0 1 0 0 0 1 0 0 0 Dimension = 3 QUESTION: 3c) /* page 88, #2.3.16 */ The space of all symmetric 3x3 matrices. SOLUTION 3c) There are as many independent elements in the basis as there are entries on-or-below the diagonal, namely n(n-1)/2 = 6 when n = 3. Here is a basis: 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 1 Dimension = 6 QUESTION: 3d) /* page 70, #2.1.7 e */ The subspace of R^infinity which consists of all arithmetic progressions, including (1/2, 3/2, 5/2, 7/2, 9/2,... ) ( 10, 7, 4, 1, -2,... ) ( 17, 18, 19, 20, 21,... ) SOLUTION 3d) A convenient basis is the "first term" and the "constant difference" of 1: ( 1, 1, 1, 1, 1, 1, ...) ( 0, 1, 2, 3, 4, 5, ...) Dimension = 2. QUESTION: 4) Let I denote the square nxn identity matrix, and let J denote the nxn matrix which has "1" is each of its n^2 components: 1 1 1 ... 1 1 1 1 ... 1 J = 1 1 1 ... 1 ..... ... . 1 1 1 ... 1 Let b be a scalar. In terms of b and n, determine the value of another scalar, c, such that (I + bJ) is the inverse of (I + cJ). SOLUTION 4) (I + bJ) (I + cJ) = I + bJ + cJ + bcJ^2 So we evaluate J^2, and notice that J^2 = nJ. Substituting this into the prior equation gives = I +(b + c + bcn) J If this is to be equal to I, all of the off-diagonal entries must be zero, so that b + c + bcn = 0 whence c = -b/(1 + bn)