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Math 415 - Assignment 4 Solutions
Problems: 2.4.2, 2.4.3 (c), 2.4.6, 2.4.8 (a0, 2.4.8 (b), 2.4.11, 2.4.15, 2.5.3, 2.5.4, 2.5.21 (c), 2.5.23, 2.6.2, 2.6.
Problem 2.4. (a) No. R^3 is 3 dimensional and so any basis of it needs exactly 3 vectors
(b) Check linear independence:
Since there are 3 pivots, these vectors are linearly independent and so form a basis of R^3.
(c) Check linear independence:
Here there are only two pivots, so the vectors are linearly dependent, and so not a basis. (d) No: more than 3 vectors cannot form a basis for a space of dimension 3
Problem 2.4.3 (c) Look at the hyperplane equation x + 2y + z − w = 0 as a single linear equation to be solved. We treat x as the basic variable and y, z, w as free and solve: x = − 2 y − z + w. Thus the most general vector in the hyperplane is
x y z w
− 2 y − z + w y z w
=^ y
+^ z
+^ w
The three vectors on the right hand side are a basis for the hyperplane.
Problem 2.4. (a) as in the last problem, let us find a basis: solve x − 2 y − 4 z = 0 for x to give x = 2y + 4z and then write the most general vector in the plane as
x y z
2 y + 4z y z
(^) = y
(^) + z
We see that a basis consists of two vectors (i.e. the plane has dimension 2) and the basis here is one of the one’s listed. To handle the other one, the two vectors involved must be inearly independent and their span must lie in the plane. For linear independence we check:
Since there are two pivots, we have independence. Now we just have to check the span. The span of these vectors consists of all vectors of the form
a
(^) + b
2 a −a + 2b a − b
Since (2a) − 2(−a + 2b) − 4(a − b) = 0, this vector satisfies x − 2 y − 4 z = 0 for all a and b, i.e. for all vectors in the span. (b) We need to find scalars c 1 and c 2 such that
c 1
(^) + c 2
c 1 c 2
and scalars c 3 and c 4 such that
c 3
(^) + c 4
c 3 c 4
We can do these two together through the following double elimination:
1 2 0 − 1 2 1 2 −^1
1 2 0 − 1 2 0 0
We conclude that c 2 = − 1 , c 4 = 2, c 1 = 12 − 12 c 2 = 1, c 3 = 0 − 12 c 4 = −1. Here is a check:
(c) If v 1 and v 2 are the vectors in the first basis, then we can form new bases in many ways. For example, 2v 1 and 3v 2 constitute a new basis. So are v 1 and v 1 − v 2. Any two linear combinations of v 1 and v 2 that are linearly independent will make up a new basis.
Problem 2.4.8 (a) First solve( Ax = 0: 1 2 − 1 1 3 0 2 − 1
so y = 56 z − 23 w and x = −2( 56 z − 23 w) + z − w = − 23 z + 13 w and therefore the solutions are given by
x y z w
− 23 z + 13 w 5 6 z^ −^
2 3 w z w
=^ z
5 6 1 0
+^ w
1 3 − (^23) 0 1
The latter two vectors form a basis for the subspace and so the subspace has dimension 2
Problem 2.4.8 (b) We proceed as in the last problem. Since p(x) = ax^2 + bx + c, the condition p(1) = 0 is a + b + c = 0. This is a linear equation in abc space. Let’s solve it. Treat a as a basic variable and b and c as free variables. Thus a = −b − c. Hence the most general p(x) satisfying p(1) = 0 is p(x) = (−b − c)x^2 + bx + c = b(x − x^2 ) + c(1 − x^2 ) This shows that our space of p(x) functions is spanned by p 1 (x) = x − x^2 and p 2 (x) = 1 − x^2. All we need to do is to show that p 1 and p 2 are linearly independent. If so, they form a basis for our set of functions and so the set has dimension 2. For linear independence we consider what linear combinations can equal the zero function: b(x − x^2 ) + c(1 − x^2 ) = 0 ⇒ when x = 0 we get c = 0 and when x = −1 we get − 2 b = 0. Thus the only linear combination giving the zero function is the zero combination. We are done
Problem 2.4. (a) We know that P (3)^ has dimension 4. Since we have been given 4 functions, they will form a basis if they are linearly independent. To show this we must show that c 1 (1) + c 2 (1 − t) + c 3 (1 − t)^2 + c 4 (1 − t)^3 = 0 implies that all of the ci’s are zero. Expanding and collecting terms, we obtain c 1 + c 2 (1 − t) + c 3 (1 − 2 t + t^2 ) + c 4 (1 − 3 t + 3t^2 − t^3 ) = 0 ⇒ (c 1 + c 2 + c 3 + c 4 ) + (−c 2 − 2 c 3 − 3 c 4 )t + (c 3 + 3c 4 )t^2 − c 4 t^3 = 0 ⇒ c 1 + c 2 + c 3 + c 4 = 0, −c 2 − 2 c 3 − 3 c 4 = 0, c 3 + 3c 4 = 0, −c 4 = 0 The last step is a consequence of the fact that the functions 1, t, t^2 , t^3 are linearly independent and so the coefficients in this linear combination of them must vanish. We have reduced to problem to that of solving 4 equations in 4 unknowns for the ci’s. A careful examination of these shows that all the ci’s are zero. (b) Here we want ci’s such that c 1 + c 2 (1 − t) + c 3 (1 − 2 t + t^2 ) + c 4 (1 − 3 t + 3t^2 − t^3 ) = 1 + t^3 ⇔ (c 1 + c 2 + c 3 + c 4 ) + (−c 2 − 2 c 3 − 3 c 4 )t + (c 3 + 3c 4 )t^2 − c 4 t^3 = 1 + t^3 ⇔ c 1 + c 2 + c 3 + c 4 = 1, −c 2 − 2 c 3 − 3 c 4 = 0, c 3 + 3c 4 = 0, −c 4 = 1 The solution of these equations (they are already in elimination form) is c 4 = − 1 , c 3 = 3, c 2 = − 3 , c 1 = 2. Here is a check: 2 − 3(1 − t) + 3(1 − 2 t + t^2 ) − (1 − 3 t + 3t^2 − t^3 ) = 1 + t^3
z =
x y z
(^) = z
The general solution of the initial system is then
x = x∗^ + z =
(^) + z
1 + z 2 + z 3 + z
. Here is a check:
Ax =
1 + z 2 + z 3 + z
Problems 2.5.21 (c)
A =
We conclude that the first two columns of A are a basis for the range of A since these are the columns corresponding to the pivots. For the range we note that y = − 3 z + 2w, x = −(− 3 z + 2w) − 2 z − w = z − 3 w and so
z =
x y z w
z − 3 w − 3 z + 2w z w
=^ z
+^ w
and the two vectors on the right are a basis for the kernel of A. Now we do the same thing with AT^ :
AT^ =
We conclude that the first two columns of AT^ are a basis for the range of AT^ , i.e. the corange of A, since these are the columns corresponding to the pivots. For the range we note that y = z, x = −y − 2 z = − 3 z and so
z =
x y z
− 3 z z z
(^) = z
and so the vector
)T
is a basis for the kernel of AT^ , i.e. the cokernel of A Problem 2.5. First we do Gaussian elimination on
A =
Since the pivots correspond to columns 1 and 3, a basis for the range of A consists of
)T
and
)T
. To write the columns of A as linear combos of these, we just have to do this for column 2, so we do the following reduction:
and this implies that
)T
)T
. For the corange we do
AT^ =
so a basis of the corange consists of
)T
and
)T
. To write the columns of AT^ as linear combos of these, we just have to do this for column 3, so we do the following reduction:
and this implies that
)T
= c 1
)T
)T
where 4c 2 = 1, i.e. c 2 = 14 and c 1 = − 3 − 2 c 2 = − 72. Check:
− (^72)
)T
)T
)T
Problem 2.6.
(a) A =
corresponds to the figure at the top of the page.
(b) Since the graph is connected, the kernel is spanned by
)T
For the cokernel we need to reduce
AT^ =
and by back substitution we find that cokernel A = { 0 }, i.e. no circuits (obvious from the figure). (c) None!
Problem 2.6. The incident matrix A for such a graph would be n × n. We know (Proposition 2.51) that if the graph is connected, then the kernel of A is one dimensional and so A has rank n − 1. This means that the rank of AT^ is n − 1 and so the cokernel of A is one dimensional, i.e. there is one circuit. If the graph is not connected, focus on a connected part of it and use a part of this subgraph with the same number of vertices and edges and apply the argument above.
