Solution Set 4, Fall 12

Transcription

1 Solution Set 4, Fall Do Problem 7 from 3.6. Solution. Since the matrix is invertible, we know the nullspace contains only the zero vector, hence there does not exist a basis for this subspace. Also, since the matrix is invertible and 3 by 3, it has rank 3. Hence we know its column space and its row space are all of R 3. A basis for this space is, for example, the standard basis: (1, 0, 0), (0, 1, 0), (0, 0, 1), but taking the 3 columns or the 3 rows of A would also work. And because A is invertible, its transpose is also invertible (check this), so the left nullspace is also just the zero vector. What about B? Since each column of A is repeated, but those columns are independent, the column space of B is the same as A, that is, R 3, and a basis is the same as above. If the column space has dimension 3, then the rank of B is 3, hence the row space is R 3 and a basis for this could be the 3 rows of B. But now the nullspace of B has dimension (number of columns)-(rank)=6-3=3. If we work out the row reduced echelon form of B, we will obtain [I 3 3 I 3 3 ]. From this we have columns 4,5, 6 as free columns, and we can obtain a basis for the nullspace, which would be the 3 vectors (-1,0,0,1,0,0),(0,-1,0,0,1,0) and (0,0,-1,0,0,1). As for the left nullspace, if we find the RREF of B T, we will find [I ] T, so that there are no free columns, and the left nullspace is just the zero vector, as expected from the formula for the dimension: (number of columns)-(rank)=3-3=0. 2. Do Problem 8 from 3.6. Solution. For A: the row and column space have dimension 3, the nullspace has dimension 5-3=2 and the left nullspace has dimension 3-3=0. For B: the row and column space have dimension 3, the nullspace has dimension 6-3=3 and the left nullspace has dimension 5-3=2. For C: the row and column space have dimension 0, the nullspace has dimension 2-0=2 and the left nullspace has dimension 3-0=3. 3. Do Problem 14 from

2 Solution. We have an LU decomposition of A: A = LU. We can see from U that the rank of A is 3. A basis for the row space is, for example, the 3 rows of U. A basis for the column space will be the pivot columns of A, that is, columns 1,2 and 3 (don t calculate them) - but the 3 columns of L would also work, as would the standard basis, since we know A has full column rank. There is only one free variable, x 4, and we see that a basis for the nullspace will be the vector (0, 1, 2, 1). To find the left nullspace, we simply note that A T has 3 columns and rank 3, hence the left nullspace is just the 0 vector. 4. Do Problem 21 from 3.6. Solution. (a) Vectors u and w span the column space of A. (b) Vectors v and z span the row space of A. (c) The rank is less than 2 if u and w are colinear (or dependent) or if v and z are colinear (or dependent). (d) A = = We swap rows 1 and 3, then 2 and 3, and obtain the RREF of A: the rank is 2, as we expect from (c). 5. Do Problem 23 from 3.6. Solution. From the decomposition we are given, we see that the column space will be spanned by the column of the first matrix on the right hand side, and those 2 columns are independent, hence a basis for the column space of A is the vectors (1, 4, 2) and (2,5,7). Similarly, the row space of A will be spanned by the rows of the second matrix on the right hand side. (To see this, take say the first row of A: each of its entries is the dot product of the first row of the first matrix, (1,2), with each column of the second matrix. Then convince yourself that in fact the first row of A is the sum of 1 times the first row of the second matrix plus 2 times the second row of the second matrix.) And since the 2 rows of the second matrix are independent, they are a basis for the row space of A. Hence now we know that the rank cannot be more than the number of columns of the first matrix, or the number of rows of the second matrix, here 2. But A is 3 by 3, hence it cannot be invertible. 2

3 6. Do Problem 26 from 3.6. Solution. Generalizing the previous exercise, we see that: If AB = C, the rows of C are combinations of the rows of B. So the rank of C is not greater than the rank of B. Since B T A T = C T, the rank of C is also not greater than the rank of A T (or A). 7. Do Problem 29 from 3.6. Solution. Let s try to find a 3 by 3 matrix with 5 ones and 4 zeros which could correspond to a completed tic-tac-toe game. Obviously, there cannot be ones in all 4 corners. We could have 3 ones in corners - the only possible configuration then is the following (up to rotations): , but this matrix has rank 3. If there are ones in exactly 2 corners, then one possible configuration is the following, corresponding to a rank 2 matrix: This matrix corresponds to the following game, where the numbers indicate the n th move (so odds are X s and evens are O s). We can see no one passes a winning move: Do Problem 3 from 8.2. Solution A = 1 0 1, U = 0 1 1, and the 2 non-zero rows of U correspond to the tree with edges from node 1 to 2 and 2 to 3. 3

4 9. Do Problem 8 from 8.2. Solution. We have the following A (slightly different answers were possible) and its echelon form U: A = , U = , from which we find a solution x = (1, 1, 1, 1) to Ax = 0. We also have A T and its echelon form V : A T = , V = , from which we find two independent solutions y = (1, 1, 1, 1, 0) and (0, 1, 1, 0, 1) to A T y = 0. Of course if your A was different from the above, your answers might have differed too. 10. Do Problem 2 from 4.1. Solution. See Figure 1. The point was for you to figure out the dimension of each space and where the arrows go. The nullspace is the zero vector, which means that the nullspace part of any vector x in R 2 is x n = 0. 4

5 Figure 1: Redraw of Figure 4.3. row space dim 2 column space dim 2 x=x_r+x_n Ax=Ax_r=b b R 2 Ax_n=0 nullspace dim 2-2=0 left nullspace dim 3-2=1 R 3 5