MH1200 Final 2014/2015

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1 MH200 Final 204/205 November 22, 204 QUESTION. (20 marks) Let where a R. A = 2 3 4, B = 2 3 4, 3 6 a For what values of a is A singular? 2. What is the minimum value of the rank of A over all a R? 3. Let b = 4. 0 For what values of a does Ax = b have infinitely many solutions? 4. Find a lower triangular matrix L and an upper triangular matrix U such that B = LU. Solution. We do Gaussian elimination on A and go ahead and record the elementary matrices which will be needed in part (4) a 0 3 a 3 This is accomplished by adding 2 times row to row 2 and adding 3 times row to row 3. The corresponding elementary matrices are E = 2 0, E 2 =

2 Continuing, we next add 3 times row 2 to row 3 This corresponds to the elementary matrix 0 0 E 3 = and brings our matrix in row echelon form a a 9 Now we can see that if a = 9 the matrix will have an all zero row. From here we can say many things the rank will be 2 and so won t be full, it has a free column thus Ax = 0 has infinitely solutions, or it has zero determinant to conclude that it is singular. If a 9 then the matrix has 3 pivots thus it will be full rank, will have nonzero determinant, does not have a free column, etc. and so is invertible. 2. No matter what a is there will be two nonzero pivots, thus the minimum value of the rank is Now we do the same row operations we did above on b which gives If a 9 then we saw above that the matrix is invertible, thus Ax = b will not have infinitely many solutions. If a = 9 then the system is inconsistent as the third coordinate of the reduced form of b is nonzero, whereas the third row of the reduced form of A is all zero. Thus for no values of a does Ax = b have infinitely many solutions. 4. When the (3, 3) entry of A is set to 0 (so that it equals B) the reduced form above becomes U = We thus have E 3 E 2 E A = U and so A = E E2 E3 U. The product E E2 E3 will be lower triangular and will form our L L = = It can be checked that B = LU. 2

3 QUESTION 2. (20 marks). Give an example of a 3-by-3 matrix with every entry either 0 or whose determinant is 2. Show that this is the largest possible value of the determinant of a 3-by-3 matrix with every entry either 0 or. 2. Let A be a 3-by-3 matrix where every entry is either 0 or and det(a) 0. What is the largest number of s that A can have? Give an example, and show that it is optimal. 3. Let A be a 3-by-3 matrix that is skew-symmetric. This means that A(i, j) = A(j, i) for all i, j 3, or equivalently A = A T. Show that det(a) = Let A, B be n-by-n matrices. Let 0 n n be the all zero matrix of size n. Show that ([ ]) A 0n n det = det(a) det(b) 0 n n B Solution. Consider a 3-by-3 matrix a b c A = d e f. g h i Using the Big Formula, we see the determinant of A is det(a) = aei + bfg + cdh afh bdi ceg. () Each term in this sum is either 0 or. If all of a, b, c,..., i are equal to, then the matrix has identical rows and the determinant is 0. Otherwise, at least one of a, b, c,..., i is equal to zero. Every letter appears twice in (), once positively and once negatively. Thus if some entry is zero, then at most 4 terms in () will be nonzero, two occurring positively and two negatively. Now as each product of three letters is either 0 or, the negative terms cannot increase the sum, thus the sum is at most 2. This reasoning also tells us how to construct an example. Some entry must be zero, let s choose i. Then the determinant becomes bfg + cdh aft ceg. Now we have to choose b = f = g = c = d = h = and can set the remaining variables to zero a = e = 0. This gives the matrix We can double check that the determinant is indeed 2. 3

4 2. Let s start with as many ones as possible, the all ones matrix. A =. This has zero determinant because two rows are the same. Now we further see if we only change one entry to a zero, the determinant will still be zero because there will be two rows that are the same, both all one rows. Thus we have to change at least two entries to zeros. Can we achieve a nonzero determinant with a matrix that has two zeros? Let s try A = 0. 0 Now we can check the determinant of this matrix is so it works. Thus there is a 3-by-3 zero/one matrix with 7 ones and nonzero determinant, and this is the largest possible. 3. We have det(a) = det(a T ) = det( A) = ( ) 3 det(a). This means det(a) = det(a) hence det(a) = As done with some theorems about determinant in lecture, we can prove this one by starting with simple matrices and building are way up. Let s give the 2n-by-2n matrix with diagonal blocks of A and B in the question a name, call it C. First, suppose that both A and B are upper triangular then det(a) is just the product of the diagonal entries of A and similarly for B. If both A, B are upper triangular, then C is upper triangular as well. Thus det(c) will be the product of its diagonal entries, which is equal to det(a) det(b). The same argument works if A, B are lower triangular. We have thus answered the question in two important special cases. Now consider the case were both A, B are permutation matrices. In this case, C is also a permutation matrix thus its determinant will be + or. Moreover the row swaps to change C to the identity are simply the row swaps to change A to I n followed by the row swaps to change B to I n. Thus the total number of row swaps will be odd iff exactly one of det(a), det(b) is. This shows det(c) = det(a) det(b). Having shown these two special cases, we can solve the general case by using LU decomposition with the product formula for determinant. Let P A A = L A U A and P B B = L B U B, where the P s are permutation matrices, the L s are lower triangular, and the U s are upper triangular. Note that this means A = PA T L AU A and so det(a) = det(pa T ) det(l A) det(u A ), and similarly for det(b). We can use these decompositions to write the matrix we are interested in as a product of determinants we can compute. [ ] [ ] [ ] [ ] A 0n n P T = A 0 n n LA 0 n n UA 0 n n 0 n n B 0 n n PB T 0 n n L B 0 n n U B 4

5 By the product formula, together with our arguments above, the determinant on the right is det(p T A ) det(l A ) det(r A ) det(p T B ) det(l B ) det(r B ) = det(a) det(b). 5

6 QUESTION 3. (30 marks) Suppose that Ax = b reduces by the usual row operations to Ux = c where x Ux = x 2 b x = b 2 b = c b x 3 2b 2 + b 4. Give a condition on b for Ax = b to have a solution. 2. Give a basis for the column space of A. 3. Is the column space of A the same as that for U? Justify your answer. 4. Give a basis for the nullspace of A. 5. Give a basis for the row space of A. 6. If C is a 3-by-4 matrix of rank, what are the possible values of the rank of A + C? Solution:. The system will be consistent if and only if b 3 2b 2 + b = For this question, we have to work backwards to find A. As the second coordinate of c is b 2 b the first operation of Gaussian elimination must have been to add times the first row of A to the second. Let s see what happened to the third row. If we add a times the first row to the third row to eliminate the (3, ) entry of A then the third coordinate of the right hand side becomes b 3 + ab. In the next step we add a multiple d times the second row to the third row, which changes the third coordinate of the right hand side to b 3 + ab + d(b 2 b ). Simplifying, this is b 3 + db 2 + (a d)b. Equating this to c we find d = 2 and a =. Thus the steps in Gaussian elimination were: ) add times the first row to the second, 2) add times the first row to the third, 3) add -2 times the second row to the third. Reversing these operations we find A = With all that done, we note that the first and third columns of U are pivot columns, thus the first and third columns of A will be a basis for the column space {(2, 2, 2), (6, 0, 4)} 6

7 3. The column space of A is not the same as that of U. For example the column space of A contains (2, 2, 2) whereas every vector in the column space of U will have third coordinate zero. 4. Elementary row operations do not change the nullspace, thus we can find a basis for the row space using U. The free columns are columns 2, 4. The special solutions corresponding to these free columns will be a basis for the row space. The special solution corresponding to column 4 is (, 0,, ) and the special solution corresponding to column 2 is ( 2,, 0, 0). Thus a basis for the nullspace is {( 2,, 0, 0), (, 0,, )}. 5. The row space of A will be the same as that for U, thus a basis for the row space of A is {[2, 4, 6, 8], [0, 0, 4, 4]}. 6. The possible values of the rank of A + C are 3, 2,. We know that A = EU for an invertible matrix E (a product of elementary matrices). If a matrix D is rank one, then ED will be rank one as well, as elementary row operations do not change the row space, and for the same reason U + D will have the same rank as EU + ED = A + ED which is A plus a rank one matrix. Thus we can think about the rank of A + C simply by studying U + D for a rank one matrix D. We can see that each of, 2, 3 are achievable as the rank of U + D by letting D = 4E 2,3 4E 24, D = E, D = E 44 respectively, where E ij is the 3-by-4 matrix with a in position (i, j) and zeros elsewhere. Finally, these are the only values that are achievable. We showed in lecture 3B that rank(x+ Y ) rank(x)+rank(y ) for any two matrices X, Y. This means rank(u +D) rank(u)+ rank(d) 3. Also rank(u) rank( D) + rank(u + D) meaning that rank(u + D). 7

8 QUESTION 4. (30 marks). Consider the subset S of R 3 defined as S = {(x, y, z) : x + y + z }. Is S a subspace? Justify your answer. 2. Consider the subset S of R 3 of all vectors (x, y, z) such that x+y+z = 0 OR x+2y+3z = 0. Is S a subspace? Justify your answer. 3. Show that if S and T are subspaces of a vector space V, then U = {s + t : s S, t T } is a subspace as well. 4. Let A, B be two matrices with the same number of columns. Describe the nullspace of the matrix [ ] A B in terms of the nullspaces of A and B. 5. Let S be the set of all 3-by-3 matrices A that are skew-symmetric, that is that satisfy A = A T. Show that S is a subspace of the vector space of 3-by-3 matrices and determine its dimension. 6. Let S be the subspace of 3-by-3 skew-symmetric matrices as in part (5). Let T be the subspace of all 3-by-3 matrices that are symmetric. Determine the dimension of the intersection of T and S and the dimension of the span of the union of T and S. Solution. It is not a subspace, (, 0, 0), (0,, 0) S but (, 0, 0) + (0,, 0) = (,, 0) S. 2. It is not a subspace, (, 0, ) S (satisfies the first equation) and (2,, 0) S but (, 0, ) + (2,, 0) = (3,, ) S. 3. We check the three conditions. As S, T are subspaces, 0 S and 0 T thus 0+0 = 0 U. If s + t U then c(s + t) = cs + ct U since cs S, ct T as they are subspaces. Finally, if s + t, s 2 + t 2 U then s + t + s 2 + t 2 = (s + s 2 ) + (t + t 2 ) U since s + s 2 S, t + t 2 T as they are subspaces. 4. If a vector x is in the nullspace of this matrix then Ax = 0 AND Bx = 0. Thus the nullspace of this matrix is the intersection of the nullspaces of A and B. 5. We check the three conditions of being a subspace. The all zero matrix = 0 T 3 3 thus it is in S. If A S then ca = c( A T ) = c(a T ) so ca S. 8

9 Finally, if A, B S then A + B = A T B T = (A + B) T thus A + B S. Now we find a basis. The subspace S is defined by a system of homogenous linear equations in the matrix entries. The general solution is 0 a b a 0 c. (2) b c 0 By separation of free parameters, we obtain a basis for S , 0 0 0, These matrices clearly span any matrix of the form in (2) and are linearly independent as each one is nonzero in an entry where all the others are zero. Thus they form a basis and the dimension of S is If a matrix is both symmetric and skew-symmetric then A = A T = A, so A is the all zero matrix. Thus the dimension of S T is zero. We claim that any 3-by-3 matrix is in the span of the union of S and T. Take any 3-by-3 matrix A. Then A = 2 (A + AT ) + 2 (A AT ). As A + A T is symmetric and A A T is skew-symmetric, their sum will be in the span of the union of S and T. This shows that the span of S T is equal to M(3, 3) the set of all 3-by-3 matrices and so has dimension 9. 9

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