SolutionSampleTest2MA3160F2012

7/30/2019 SolutionSampleTest2MA3160F2012

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Solution to Sample Test 2 MA 3160

Chapter 3 and Chapter 4.1,, 4.2, 4.3, 4.4,4.5

[1] Understand material Quiz 3, Chapter 3 Practice Problem

[2] From Chapter 4:

1. Decide if W={(x,y,z) R3 | x+y+z=0} is a subspace of R3

Solution: We need to show that

1) W and 2) Closure of +. We need to show that

for any

,

3) Closure of scalar multiplication: For any 1) W because (0,0,1) W. Since all elements of W are from

2) Since , , and Then

Now z2=0+0=0 Hence u+v

3)

Since , Since it satisfies all three properties of a subspace, W is asubspace.


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2. Decide if W={(x,y,z) R3 | x+y+z=1} is a subspace of R3

Solution: We will show that W is not a subspace by presenting acounter example which violates the closure of scalar multiplication.

Note that (1,1,-1) W since 1+1+(-1)=1.

But 2(1,1,-1) =(2,2,-2) since 3. Decide if is a subspace of M2,2, a space

of all 2 x2 matrices. You may assume that M2,2 is a vector space.

Solution: We need to show that

1) W2) Closure of +. We need to show that for any , 3) Closure of scalar multiplication: For any

For 1) note that W because

.

For 2) suppose that , Then and for some real numbers Now + = Wsince it is a 2 x 2 diagonal matrix ( (2,1) entry and (1,2) entry

are 0) and For 3) note that


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Now since

Hence W is a subspace.

4. Let W be the set of all points . Decide if W is asubspace of.Solution: We will find a counter example which violates the closure ofscalar multiplication and W is not a subspace of Note that (1,2) W since . But for k=(-2) R,k(1,2)=(-2)(1,2)= (-2,-4)

Hence W is not a subspace of R2

5. Decide if is a linear combination of Soluton: is a linear combination of only if: for some number a and b. (*)We will try to solve and decide if (*) has a solution.

Form augmented matrix from (*).

By elementary row operations


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This system has a solution

Hence : a linear combination of 6. Show that

Solution: Recall that two sets A and B are equal only if

Since To show that be an arbitrary elementof We need to find x and y such that Which is equivalent to solving

Since the inverse of


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Hence there is x and y such that and Hence

7. Decide if

Solution: we will check if t there are numbers a and b such that

Putting it in Augmented matrix form,

by replacing R3 with R3-R2.

We notice that the last row shows that the system is inconsistent, (no

solution: . Hence if

8. Decide if Solution: In 7, we showed that


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.In 7. We showed that Hence

9. Decide if

is a linearly independent set.Solution: To show linear indepency, we will check if (a,b,c) = (0,0,0)is the only solution.

a

Form augmented matrix

Then the system beceoomes

This system has a unique solution a=0,b=0,c=0.


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Hence

is a linearly independent set.10. Show that

a basis of a vector space

Solution: We need to show that is linearly independentand they span R2.

In number 6, we showed that spans R2Since

has a

unique solution.

Hence it is linearly independent.

Since is linearly independent and spans R2, 1 a basis of a vector space R2

11. Show that

Solution: We need to show that hasonly trivial solution (not a zero )

This equation becomes = .This gives us

This system has only trivial solution a=0, b=0. Hence we conclude

that


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12. Show that is a basis for W , the set of all2x2 diagonal matrices.

Solution: We will show that

is linearly

independent and spans Linearly independent: In number 11, we showed that 1001 is linearly independent. (you will need to put the proof in 11,here)

To show that spans W, suppose that a matrixA W. Then

.

We will show that there are numbers x and y such that

=x Since

= the problem is to check if

has a solution x andy.

Writing in a matrix form of equation,

(*)


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Since has an inverse. Hence for any a, b R, (*) has a solution

Since is linearly independent and spans

Find a coordinate of

relative to a basis S={

Solution: From 5 we have S={ 1 of10is (-3/2, -1/2) .

Reamark. : To signify the basis the coordinate uses, we put subscript

S next to the vector 13. Find the dimension of the spaces in the above problems 2,4,7.

Skip.14. Show that V= satisfies the following

properties of a vector space.

Let u,v are in V and c R. Then

a. u+v Vb.c. u+v=v+u

d. There is a special element called the zero vector in V (denote by 0) suchthatfor any u V, u+0=u and 0+u=u.

Solution. Suppose that u and v are in V. Then


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for some number andthey satisfies

a. Proof that u+v V:

Note .Then by properties ofreal number.

Hence u+v= b. Proof that cu V. Notice that

Note that Hence cu Vc. Proof that u+v=v+uNote that u+v= since + is commutative in R.=

15. (Extra credit , optional . Study only if you are comfortable withall other material in this course)

Let the set of all polynomials ofdegree 2 or less. Show tha V satisfies the following properties of avector space where

addition is defined as

and the scalar multiplication is defined as for any k R and Show that for any u,v V and k R .


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a. u+v V

b. There is a special element called the zero vector in V (denote by 0)such thatfor any u V, u+0=u and 0+u=u.

Proof: Note that two polynomials if and only if a. Proof of a. Let u V and v V. We will have to show that

u+v is well definedNote that the last relation

holds

because

b. We claim that 0 is in V and it satisfies the property of t of

the vector space.Notice that Further, let u V. Then .Since

0 is the zero vector in V.

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SolutionSampleTest2MA3160F2012