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Math 416 Homework 2. Solutions.

0. Show that if a real vector space has two elements, it must have innitely many elements.Solution:(This was, at one point, problem #1 of the assignment, but then I deleted it. So I'm

calling it problem #0 here and giving the solution, to make sure I don't break the numbering for the problems that were actually assigned.) We have proved that the zero of a vector space is unique, so if it has two elements it must have at least one nonzero element. Let us call thisx6=0. Then note thatxis also in the vector space for any2R, since the vector space is closed under scalar multiplication. It is easy to see that if we can show thatfxj2Rgis innite then we are done. But it is not obvious at this stage that it must be so, e.g. it could be thatx=xfor all, or something like that. In fact, we will prove something slightly stronger:

Claim.Ifx6=0and6=thenx6=x.

If the claim is true, then we are done, because the setfxj2Rghas as many dierent elements asR, which of course is innite. Thus we need only prove the claim. Let us assume thatx6=0and6=. We will use a proof by contradiction, so let us assume thatx=x. Add the additive inverse ofxto both sides, and we obtain x+ ((x)) =x+ ((x)); x+ ((x)) =0; xx=0; ()x=0: The rst step is the denition of additive inverse, the second step uses Theorem 1.3 (i.e. (x) = ()x), and the third step uses one of the distributive axioms. Now, if6= 0, then1=()is a real number, so multiplying both sides of this equation by this number gives x=1()x=10=0; where the last equality is again using Theorem 1.3. But this says thatx=0, which is a

contradiction! Therefore the claim is true and we are done.1.(q.v. Freidberg, Insel, Spence 1.4.8).Show thatPn(R), the set of all polynomials with

real coecients and degree less than or equal ton, is generated by the setf1;x;x2;:::;xng.Solution:The denition of a polynomial is that it is a linear combination of monomials. If

the polynomial is of degreenor fewer, then it can be written as a linear combination of powers less than or equal ton.

2.(q.v. Freidberg, Insel, Spence 1.4.11).LetVbe a vector space overF, andx2V. Prove

that Span(fxg) is the setfax:a2Fg. Interpret this result geometrically inR3.Solution:By the denition ofSpan(),y2Span(fxg)if and only if it can be written as a

linear combination of vectors infxg| but since there is only one vector infxg, then the linear combinations are of the formxwith2F. If we think ofR3, then there are two cases: eitherx=0orx6=0. If the former, we see that the span is just the zero vector itself, since0=0by Theorem 1.3. If the latter, then x= (a;b;c)with not all ofa;b;czero, and thusx= (a;b;c)is the line inR3spanned

by the vector(a;b;c).3.(q.v. Freidberg, Insel, Spence 1.4.13).Show that ifS1;S2are subsets of a vector space

VandS1S2, then Span(S1)Span(S2). Deduce from this that if Span(S1) =V, then Span(S2) =V.Solution:Letx2Span(S1). This means that for some vectorsv1;v2;:::;vn2S1, we have x=a1v1+a2v2++anvn: But notice thatvi2S2for alli, sinceS1S1. Thereforexis a linear combination of vectors inS2, sox2Span(S2). To prove the second half, notice that sinceVis a vector space,Span(S2)V, and by the

previous part,Span(S2)Span(S1) =V, so we must haveSpan(S2) =V.4.(q.v. Freidberg, Insel, Spence 1.4.14).Show that ifS1;S2are two subsets of a vector

spaceV, then Span(S1[S2) = Span(S1) + Span(S2).Solution:To prove this, we need to think about what the two sides of this equation mean.

Recall that ifW1;W2are subsets of a vector spaceV, thenW1+W2is the subset ofVgiven by all sums of the formx1+x2, wherexi2Wi. Thus in words, the equation says that any linear combination of vectors lying in the union of S

1andS2can be written as a sum of two vectors, one of which is a combination of vectors in

1, and the other a combination of vectors inS2. (This is a correct, and short, proof | but

notice that short proofs are very delicate, if you change any word in that sentence it breaks!) Using a more equation-based argument: letx2Span(S1[S2). This means that we can write x=a1u1+a2u2++anun+b1v1+b2v2++bmvm; whereui2S1;vi2S2. But then we can write x= (a1u1+a2u2++anun) + (b1v1+b2v2++bmvm); and clearly this lies inSpan(S1) + Span(S2).

5. Write down the solution set of the system

1+ 2x2+ 3x3= 7;

2x1+ 3x26x3= 3;

9x1+x23x3=1:Solution:We do the row reduction:

0 @1 2 3 6

2 36 3

9 1311

A ;0 @1 2 3 6 01129

01730551

A ;0 @1 2 3 6

0 1 12 9

01730551

A 0 @1 2 3 6

0 1 12 9

0 0 174 981

A ;0 @1 2 3 6

0 1 12 9

0 0 1 4987
1 A ;0 @1 2 3 6 0 1 0 6529

0 0 14987

1 A ;0 @1 0 0529

0 1 06529

0 0 14987

1 A

Thus we have the (unique!) solution

x 1=529 ; x2=6529 ; x3=4987 Not the cleanest computation around, but it is what it is. Notice something strange that happened with the fractions: we got a few dierent denominators, but notice that87 = 329.

Is this

1a coincidence???!?!?!?!?!?!?!?!?!?!??!6. Consider the linear system of 4 equations in 4 unknowns written as follows:

42x1+x2= 0;

Writing these out gives

1=x4; x2=x4; x3=x4;

so that all of the values must be equal. We can also represent this in matrix form 0 B B@x 1 x 2 x 3 x 41
C CA=0 B B@x 1 x 1 x 1 x 11 C

CA=x10

B B@1 1 1 11 C CA:7. Consider the previous problem, but now the right-hand sides of the equations are all 2. Write

Claim.Ifx6=0and6=thenx6=x.

2.(q.v. Freidberg, Insel, Spence 1.4.11).LetVbe a vector space overF, andx2V. Prove

1andS2can be written as a sum of two vectors, one of which is a combination of vectors in

1, and the other a combination of vectors inS2. (This is a correct, and short, proof | but

5. Write down the solution set of the system

1+ 2x2+ 3x3= 7;

2x1+ 3x26x3= 3;

9x1+x23x3=1:Solution:We do the row reduction:

2 36 3

9 1311

01730551

0 1 12 9

01730551

0 1 12 9

0 0 174 981

0 1 12 9

0 0 14987

0 1 06529

0 0 14987

Thus we have the (unique!) solution

Is this

1a coincidence???!?!?!?!?!?!?!?!?!?!??!6. Consider the linear system of 4 equations in 4 unknowns written as follows:

42x1+x2= 0;

12x2+x3= 0;

22x3+x4= 0;

32x4+x1= 0:

B@2 1 0 1

12 1 0

0 12 1

1 0 121

B@12 1 0

2 1 0 1

0 12 1

1 0 121

B@12 1 0

03 2 1

0 12 1

0 2 021

B@12 1 0

0 12 1

03 2 1

0 2 021

B@12 1 0

0 12 1

0 04 4

0 2 021

B@12 1 0

0 12 1

0 04 4

0 0 441

B@12 1 0

0 12 1

0 0 11

0 0 441

B@12 1 0

0 12 1

0 0 11

0 0 0 01

B@12 1 0

0 1 01

0 0 11

0 0 0 01

B@1 0 12

0 1 01

0 0 11

0 0 0 01

B@1 0 01

0 1 01

0 0 11

0 0 0 01

Writing these out gives

1=x4; x2=x4; x3=x4;

CA=x10

B@2 1 0 1 2

12 1 0 2

0 12 1 2

1 0 12 21