Download Linear Algebra Homework Solutions: Rank, Column Space, Nullspace, and Dependence - Prof. P and more Assignments Linear Algebra in PDF only on Docsity!
Department of Mathematics Linear Algebra and Probability
Homework Key
Recommended Problems
Monday, April 6
Assignment:
- p. 151: 1bcd (the rank is the dimension of the column space)
- p. 163: 1 (find the column space, the nullspace, and a complete solution), 4, 8a, 13abd, 14
- p. 178: 1, 5, 9, 11, 15, 18, 21a, 23 (omit row spaces)
Page 151
- If the rank of A is the dimension of the column space, which of these rules gives a correct rule for finding the rank.
(b) The number of columns minus the total number of rows. This is incorrect. If the number of columns is less than the number of rows, this formula would give a negative value. (c) The number columns minus the number of free columns. This is correct. The dimension of the column space is the number of pivot columns, and
number of cols = number of pivot cols + number of free cols.
(d) The number of 1’s in the matrix R. This is incorrect. In one of the homework problems in the last set (number 12 on page 141), we saw that we can vary the number of 1’s in R without affecting the number of pivot columns.
Page 163
- Find the column space, the nullspace, and a complete solution if
A =
(^) b =
The reduced echelon form of the augmented matrix is
From this we see that the column space of A has basis
The special solutions, which form a basis for the nullspace, are
s 1 =
,^ s^2 =
A particular solution (with last two variables 0) is
xp =
So a complete solution is xp + as 1 + bs 2 , where a and b are scalars.
- Find the complete solution to
x y z t
The reduced echelon form of the augmented matrix is
- Decide the dependence or independence of
(a) the vectors (1, 3 , 2), (2, 1 , 3), and (3, 2 , 1). (b) the vectors (1, − 3 , 2), (2, 1 , −3), and (− 3 , 2 , 1).
(a) If we form a 3 × 3 matrix with these columns, its reduced echelon form is I. So the vectors are independent. (b) If we form a 3 × 3 matrix with these columns, its reduced echelon form is
So the first two vectors are independent, but when the third is included, the three vectors are dependent.
- Suppose v 1 , v 2 , v 3 and v 4 are vectors in R^3.
(a) These four vectors are dependent because R^3 has dimension 3. So any set of four or more vectors must be dependent. (b) The two vectors v 1 and v 2 will be dependent if they lie on the same line. (c) The vectors v 1 and (0, 0 , 0) are dependent because any set of vectors that includes 0 is dependent. For example, 0 · v 1 + 1 · 0 = 0 is a nontrivial linear combination that gives 0.
- Describe the subspace of R^3 (is it a line, a plane, or all of R^3 ?) spanned by
(a) the two vectors (1, 1 , −1) and (− 1 , − 1 , 1). These vectors lie on a line since the second is −1 times the first. (b) the three vectors (0, 1 , 1), (1, 1 , 0) and (0, 0 , 0). The first two vectors are independent. So they span a plane. Since (0, 0 , 0) is always in any subspace, it adds nothing to this plane, and the span of the three vectors is the same plane. (c) all vectors in R^3 with whole number components. The span is all of R^3. For example, the “standard basis” vectors (1, 0 , 0), (0, 1 , 0) and (0, 0 , 1) have whole number components. (d) all vectors with positive components. The span is also all of R^3. For example, the reduced echelon form of
is I.
- If v 1 ,... vn are linearly independent, the space they span has dimension n. These vectors are a basis for that space. If the vectors are the columns of an m × n matrix, then m ≥ n. If m = n, that matrix is invertible.
- Suppose v 1 ,... v 6 are six vectors in R^4.
(a) Those vectors might not span R^4. For example, they could all be the zero vector. (b) Those vectors are not linearly independent. The maximum number of indepen- dent vectors in R^4 is 4. (c) Any four of those vectors might be a basis for R^4.
21(a) Suppose the columns of a 5 × 5 matrix A are a basis for R^5. The equation Ax = 0 has only the solution x = 0 because A is invertible.
23 U comes from A by subtracting row 1 from row 3:
A =
(^) and U =
Find bases for the two column spaces. Find bases for the two nullspaces. Which spaces stay fixed in elimination. The reduced echelon form of both A and U is
R =
So the column space of both A and U has a basis that consists of its first two columns. So a basis for the column space of A is
and a basis for the column space of U is
The nullspaces are unchanged by row operations. So both matrices have the same nullspace, which is spanned by the special solution s = (1, − 1 , 1). Elimination changes the column space (although it doesn’t change its dimension). Elimination doesn’t change the nullspace.
