8/8/2019 HW1 Math 405

1/6

8/8/2019 HW1 Math 405

2/6

George Austin Starkweather

Math 405

2/4/10

2

e) No, not closed under addition. Consider x, yof the form (1, 2,3) such that x= (1,0,1) and y = (0, 1,

1). If summed, the new vector would be (1,1,2), and so 1+ 2 =2, a violation of the conditions.

5)

a)Yes, this is closed under vector addition. Take, for instance two vectors, x = (1, t1, t12

,t13) and y = (1,

t2, t22

, t23). When added together, these vectors form a vector z = (2, t1+ t2, t1

2+ t2

2,t1

3+ t2

3), which

satisfies the equation. Under scalar multiplication, let be a scalar such that x = (*1, * t1, * t12

,

*t13). Finally, 0*x = (0,0,0,0), so the vector space also contains an origin.

b) Yes this is a vector space. It is:

Closed under addition: let z(t) = x(t) + y(t). Then if 2x(0) = x(1) and 2y(0) = y(1), then 2z(0) = 2x(0) +

2y(0) = x(1) + y(1) = z(1)

Closed under scalar multiplication: For any scalar , 2(z(t)) = 2(x(t) + y(t)) = (2x(t) + (2y(t)) =

x(1) + y(1) = (x(1)+y(1)) = z(1).

Also, there exists an origin, for 0* 2x(0) = 0 = 0*x(1).

c) No. x(t) 0 whenever . Not closed under scalar multiplication, consider -1*(x(t)) 0.

d) Yes, this is a vector space. It is:

Closed under addition: z(t) = x(t) + y(t) = x(1-t) + y(1-t) = z(1-t)

Closed under scalar multiplication: *z(t) = *x(t) + *y(t) = *x(1-t) + *y(1-t) = z(1-t).

Also, there exists an origin such that 0 * x(t) = 0.

8/8/2019 HW1 Math 405

3/6

George Austin Starkweather

Math 405

2/4/10

3

Section 1.7 #s 1, 3, 4(a, b), 5(a, b), 6, 8

1.

a)Observe that . So the set as a whole is linearly independent. However, itcan be shown that any three vectors are linearly independent.

Because there are four total vectors (x, y, z, u), we can make four groups of three [].Let . Also, let be scalars.

Group 1 includes Therefore, since all equal 0, this set of vectors is linearly independent.

Group 2 includes

To solve this system of equations, we must first observe that, in the first equation, Once that is established, the second and third equations can be solved for 0.

Group 3 includes

Group 4 includes

8/8/2019 HW1 Math 405

4/6

George Austin Starkweather

Math 405

2/4/10

4

To solve this system of equations, it must first be observed that, in the first equation, . From there, it can easily be solved for the other variables as well.Thus, since all of our systems of equation solve for our scalars = 0, each system is linearly

independent.

b) Observe that Hence, we know that the system is linearlydependent. However, if we were to build 4 similar systems of equations as in part (a) above, we

would show that any three are linearly independent.

Let

.Group 1 consists of Also, let be scalars.

Therefore, since all equal 0, this set of vectors is linearly independent.

Group 2 includes

To solve this system of equations, we must first observe that, in the first equation, Once that is established, the second and third equations can be solved for 0.

Group 3 includes

Group 4 includes

8/8/2019 HW1 Math 405

5/6

George Austin Starkweather

Math 405

2/4/10

5

To solve this system of equations, it must first be observed that, in the first equation,

. From there, it can easily be solved for the other variables as well.Thus, since all of our systems of equation solve for our scalars = 0, each system is linearly

independent.

3)

Yes, if are linearly independent vectors, then in reduced row echelon form, the make theidentity matrix

. As prompted, we will create three new vectors from ,

. This gives us the matrix

. When reduced to reduced row echelon form, this still gives us

the identity matrix

. Hence, our matrix is still linearly independent.

4)

a) We want to assess under what conditions on the scalar the vectors and are linearly dependent. Through creating a matrix from these two scalars, we will havelinear dependence when their determinants are equal to 0. Hence:

Hence, the determinant will equal zero only when

,

which is the only condition by which we will have linear dependence.

b) Just as above in part (a), we will have linear dependence only when the determinant of the matrix

containing this set of equations is equal to 0. Lets create that matrix and solve for .

8/8/2019 HW1 Math 405

6/6

George Austin Starkweather

Math 405

2/4/10

6

det

=

Hence , to satisfy linear

dependence.

5)

a) Similar to problem number 4, linear dependence can be established by demonstrating that the

determinant for the system of equations is equal to 0.

By placing our two vectors into a matrix, we have . In solving for the determinant and setting

the equation equal to 0 (for linear dependence, this reduces to This becomes , satisfying the prompt.

b) We will solve this problem by setting up a system of 3 equations , all set to 0, using as ourscalars. Hence, we shall set as our two vectors: .

Using scalars, we get the following system of equations:

Solving each equation for , we get:

Hence, to establish linear dependence, each value of x must be a set scalar multiple of thecorresponding value of y.