程序代写代做代考 Math 215.01, Spring Term 1, 2021 Problem Set #6
Math 215.01, Spring Term 1, 2021 Problem Set #6
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1. Recall that P is the vector space of all polynomial functions f:R!R. LetWbethesubsetofPconsistingof those polynomials that have a nonnegative constant term (i.e. the constant term of each polynomial is greater than or equal to 0). Is W a subspace of P? Either prove or give a specic counterexample.
Denition 4.1.12: Let V be a vector space. A subspace of V is a subset W V with the following properties:
1.~0 2 W .
2. For all w~1; w~2 2 W , we have w~1 + w~2 2 W .
3. For all w~ 2 W and all m 2 R, we have m w~ 2 W .
Let W be the subset of P consisting of those polynomials that have a nonnegative constant term. Let g(x) = x2 + x+1foranyx2R. SincePisthevectorspaceofall polynomial functions f : R ! R, we have g(x) 2 P. Since g(x) has a nonnegative constant term, we have g(x) 2 W. Let’s x c 2 R with c = 1. Notice that after multiplying g(x) by c, we have cg(x) = 1(x2+x+1) =
(1x2)+(1x)+(11) = (x2)+(x)+(1). Since cg(x) contains negative constant term 1, we know cg(x)62W. Sincec2Randcg(x)62W,weknowW is not closed under scalar multiplication which indicates W is not a subspace of P.
2. Let V = R4. Write down, but do not solve, a system of four equations in three unknowns such that
1 2 6 0 7 2 Span5; 1 ;3
0 1 8 3 6 423
if and only if the system has a solution.
2x + 6y = 1
5x + y + 3z = 7 x 8y + 3z = 0 4x + 2y + 3z = 6
3. Let V be the vector space of all 2 2 matrices. Show that
2 7 1 1
= a + b
3 0
.
5 4
2 7 1 9
2 Span
1 13 0 ; :
27 1130
Since 2 Span ; , we
1 9 2 3
can x a; b 2 R such that
5 4
1 9 2 3
After linear combination, we have
a+3ba 27
= .
2a+5b 3a4b 1 9 Then we have a system of equations:
a + 3b = 2
a=7
2a + 5b = 1
3a 4b = 9.
2 3
5 4
After solving the system of equations, we have a = 7 and
27 11 30
b=3. Since =7 +(3) ,
1 9 2 3 5 4
27 1130
we have shown 2 Span 1 9
2 3
; : 5 4
4. LetF bethesetofallfunctionsf :R!R. Recallthat a function f : R ! R is called even if f (x) = f (x) for all x 2 R. Let W be the set of all even functions, i.e.
W = ff 2 F : f (x) = f (x) for all x 2 Rg:
Is W a subspace of F? Either prove or give a specic counterexample.
Denition 4.1.12: Let V be a vector space. A subspace of V is a subset W V with the following properties:
1.~0 2 W .
2. For all w~1; w~2 2 W , we have w~1 + w~2 2 W .
3. For all w~ 2 W and all m 2 R, we have m w~ 2 W .
Denition of function addition: Suppose that f : R ! R and k : R ! R are functions. Then (f +k)(a) = f(a)+ k(a) for all a 2 R.
Denition of scalar multiplication with function: Suppose that q : R ! R is a function and let m 2 R be arbitrary. The function m[q] is dened by m[q](n) = m (q(n)) for all n 2 R.
LetF bethesetofallfunctionsf :R!R. LetW bethe set of all even functions. Let ^0 : R ! R be a zero vecotr functionsuchthat^0(x)=0forallx 2R. Sincex 2Rand x 2 R, we have ^0(x) = 0 and ^0(x) = 0. Therefore we have shown ^0 is an even function which indicates ^0 2 W . Thus, we have shown W is not empty. Let function j 2 W and function g 2 W be arbitrary. According to denition of function addition, we know (g+j)(x) = g(x)+j(x) and (g+j)(x) = g(x)+j(x) for all x 2 R. Since function g 2 W and function j 2 W, we have g(x) = g(x) and j(x) = j(x) for all x 2 R. Since g(x) = g(x) and j(x) = j(x), we have g(x)+j(x) = g(x)+j(x) which indicates (g + j)(x) = (g + j)(x). Therefore, we have shownfunction(g+j)2W. Sinceg2W andj2W are arbitrary, we have shown W is closed under vector addition. Let c 2 R be arbitrary. Let function h 2 W be arbitrary. Since function h 2 W, we know h(x) = h(x) for all x 2 R. According to denition of scalar multiplication with function, we have c[h](x) = c (h(x)) and c[h](x) = c (h(x)) for all x 2 R. Since h(x) = h(x), we
know c (h(x)) = c (h(x)) which indicates c[h](x) = c[h](x). Therefore, we have shown function c[h] 2 W. Since h 2 W is arbitrary and c 2 R is arbitrary, we can conclude W is closed under scalar multiplication. Since W satises all three properties in denition 4.1.12, we can conclude W is a subspace of F
5. Use Gaussian Elimination to solve the following system:
x z=0 3x + y = 1 x + y + z = 4:
Matrix of this linear system:
By R1 $ R3, we have
1 0 1 0
3 1 0 1.
1 1 1 4
1 0 1 0 1 1 1 4
3 1 0 1=3 1 0 1.
1 1 1 4 1 0 1 0 By replacing R2 with 3R1 + R2, we have
1 1 1 4 1 1 1 4
3 1 0 1=0 4 3 13.
1 0 1 0 1 0 1 0 By replacing R3 with R1 + R3, we have
1 1 1 4 1 1 1 4
0 4 3 13=0 4 3 13.
1 0 1 0 0 1 0 4 By R2 $ R3, we have
1 1 1 4 1 1 1 4
0 4 3 13=0 1 0 4.
0 1 0 4 0 4 3 13
By replacing R3 with 4R2 + R3, we have
1 1 1 4 1 1 1 4
0 1 0 4=0 1 0 4.
0 4 3 13 0 0 3 3 By replacing R1 with R1 R2, we have
1 1 1 4 1 0 1 0
0 1 0 4=0 1 0 4.
0 0 3 3 0 0 3 3
1
By replacing R1 with R1 R3 , we have
3
1 0 1 0 1 0 0 1
0 1 0 4=0 1 0 4.
0 0 3 3 0 0 3 3 By replacing R1 with R1 (1), we have
1 0 0 1 1 0 0 1
0 1 0 4=0 1 0 4.
0 0 3 3 0 0 3 3
1
By replacing R3 with R3 , we have
3
1 0 0 1 1 0 0 1
0 1 0 4=0 1 0 4.
0 0 3 3 0 0 1 1
Then we have following solution: x = 1; y = 4; z = 1.
The solution set is
1
4 .
1
6. Give a parametric description of the solution set of the following system:
x + 2y z = 3 2x + y + w = 4 x y + z + w = 1:
1 2 103 Matrix of this linear system:2 1 0 1 4
1 1 1 1 1
101 2 5
333 212
Reduced echelon form: 0 1 3 3 3 00000
Now we have a system with the same solution set as the
orginal system:
125 x+z+w=
333
212 yzw=.
333
512 221 Thenwehavex= z wandy= + z+ w.Let
333 333 t;s 2 R be arbitrary with
512 x=ts
333
221 y=+t+s
333 z=t
w = s.
5 1 2
333 221
S=f03+t13 +s03 :t;s2Rg
001
7. Comment on working with partner(s): We completed the problem set individually and checked our work together. Although we took dierent approaches to prove or solve these problems, we had the similar results in the end. We discussed why we have dierent forms of parametric de- cription of the solution set for problem 6. In the end, We conclude our works are mostly valid.
