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Chapter 1 Exercise C


1. Solution: (a) {(x1,x2,x3)F3:x1+2x2+3x3=0} is a subspace of F3. By 1.34, to show a subset is a subspace, we just need to check Additive identity, Closed under addition and Closed under scalar multiplication.

Additive identity: it is clear that the additive identity (0,0,0) of F3 is contained in {(x1,x2,x3)F3:x1+2x2+3x3=0}.

Closed under addition: if (a1,a2,a3),(b1,b2,b3){(x1,x2,x3)F3:x1+2x2+3x3=0}, thena1+2a2+3a3=0andb1+2b2+3b3=0.Hence(a1+b1)+2(a2+b2)+3(a3+b3)=(a1+2a2+3a3)+(b1+2b2+3b3)=0,this means (a1+b1,a2+b2,a3+b3)=(a1,a2,a3)+(b1,b2,b3) is also in {(x1,x2,x3)F3:x1+2x2+3x3=0}.

Closed under scalar multiplication: if (a1,a2,a3){(x1,x2,x3)F3:x1+2x2+3x3=0}, then a1+2a2+3a3=0. For any λF, we haveλa1+2(λa2)+3(λa3)=λ(a1+2a2+3a3)=0.This means λ(a1,a2,a3)=(λa1,λa2,λa3){(x1,x2,x3)F3:x1+2x2+3x3=0}.

(b) {(x1,x2,x3)F3:x1+2x2+3x3=4} is not a subspace of F3 since (0,0,0) is not in it.

(c) {(x1,x2,x3)F3:x1x2x3=0} is not a subspace of F3 since (1,1,0) and (0,1,1) are in it, but the sum (1,2,1)=(1,1,0)+(0,1,1) is not in it.

(d) {(x1,x2,x3)F3:x1=5x3} is a subspace of F3.

Additive identity: it is clear that the additive identity (0,0,0) of F3 is contained in {(x1,x2,x3)F3:x1=5x3}.

Closed under addition: if (a1,a2,a3),(b1,b2,b3){(x1,x2,x3)F3:x1=5x3}, then a1=5a3 and b1=5b3. Hencea1+b1=5a3+5b3=5(a3+b3),this means (a1+b1,a2+b2,a3+b3)=(a1,a2,a3)+(b1,b2,b3) is also in {(x1,x2,x3)F3:x1=5x3}.

Closed under scalar multiplication: if (a1,a2,a3){(x1,x2,x3)F3:x1=5x3}, then a1=5a3. For any λF, we have λa1=λ(5a3)=5(λa3). This means λ(a1,a2,a3)=(λa1,λa2,λa3){(x1,x2,x3)F3:x1=5x3}.


2. Solution: (a) If this set is a subspace of F4, then (0,0,0,0)F4, then 0=50+b. Hence b=0. Follow the same steps of Problem 1 (d), we will see the set is a subspace of F4 if b=0.

(b) (c) and (d) is similar to Problem 3 and 4.

Now let us consider (e). Denote the set of all sequences of complex numbers with limit 0 by A.

Additive identity: it is clear that (0,0,)A.

Closed under addition: if (a1,a2,),(b1,b2,)A, thenlimnan=0andlimnbn=0.It is easy to seelimn(an+bn)=limnan+limnbn=0+0=0.This means (a1+b1,a2+b2,)=(a1,a2,)+(b1,b2,)A.

Closed under scalar multiplication: if (a1,a2,)A, thenlimnan=0.For any λC, it is easy to seelimn(λan)=λlimnan=λ0=0.This means λ(a1,a2,)=(λa1,λa2,)A.


3. Solution: Denote the set of differentiable real-valued functions f on the interval (4,4) such that f(1)=3f(2) by V.

Additive identity: it is clear that the constant function f0 is contained in V.

Closed under addition: if f,gV, then f and g are differentiable real-valued functions. So is f+g. Moreover,(f+g)(1)=f(1)+g(1)=3f(2)+3g(2)=3(f(2)+g(2))=3(f+g)(2).This concludes V is closed under addition.

Closed under scalar multiplication: if fV, for any λR, then f is differentiable real-valued functions. So is λf. Moreover,(λf)(1)=λf(1)=λ(3f)(2)=3(λf)(2).This deduces V is closed under scalar multiplication.


4. Solution: Denote the set of continuous real-valued functions f on the interval [0,1] such that 01f=b by Vb.

If Vb is a subspace of R[0,1], then for any fVb, we have 01f=b. Because Vb is a subspace of Rn, it follows that kfVb for any kR. Henceb=01(kf)=k01f=kb,for all kR,this happens if and only if b=0.

Now if b=0, then for any f,gV0 and λR. We have that01(f+g)=01f+01g=0+0=0and f+g is continuous real-valued functions since f and g are. This deduces f+gV0, i.e. V0 is closed under addition. Similarly,01(λf)=λ01f=k0=0and λf is continuous real-valued functions since f is. This implies λfV0, i.e. V0 is closed under scalar multiplication. On the other hand, the constant function f0V0, which is also the additive identity in R[0,1]. Hence V0 is a subspace of Rn by 1.34.


5. Solution: Note that we consider complex vector space, so if R2 is a subspace of the complex vector space C2, theni(1,1)=(i,i)R2,we get a contradiction. Hence R2 is not a subspace of the complex vector space C2.


6. Solution: (a) Because a3=b3 if and only if a=b in R, hence {(a,b,c)R3:a3=b3}={(a,b,c)R3:a=b}is obviously a subspace of R3 by the similar arguments in Problem 1 and Problem 2.

(b) Note thatx=(1,1+3i2,0){(a,b,c)C3:a3=b3}andy=(1,13i2,0){(a,b,c)C3:a3=b3}.However,x+y=(2,1,0){(a,b,c)C3:a3=b3}.This implies {(a,b,c)C3:a3=b3} is not closed under addition, hence not a subspace of C3.


7. Solution: Denote {(x,y)R2:x,yZ} by U, then U is not empty. If (x1,y1)U and (x2,y2)U, then x1,x2,y1,y2Z. Hence x1+x2 and y1+y2 are integers. This means (x1+x2,y1+y2)=(x1,y1)+(x2,y2)U, i.e. U is closed under addition. Similarly, since (x1,y1)U, it follows that U is closed under additive inverses. However, U is not closed under scalar multiplication since (1,1)U while 12(1,1)U. Hence U is not a subspace of R2.


8. Solution: Denote {(x,y)R2:x=0 or y=0} by U, then U is not empty. If (x,0)U, then for any λR, we haveλ(x,0)=(λx,0)U.Similarly, λ(0,y)U, hence U is closed under scalar multiplication. However, (1,0),(0,1)U while (1,1)=(1,0)+(0,1)U. This implies U is not closed under addition, hence not a subspace of R2.


9. Solution: Denote the set of periodic functions from R to R by S. Then S is not a subspace of RR. Otherwise, we have h(x)=sin2x+cosx since both f(x)=sin2x and g(x)=cosx are periodic functions from R to R. Assume there exists a positive number p such that h(x)=h(x+p) for all xR, then 1=h(0)=h(p)=h(p). It is equivalent to 1=cosp+sin2p=cospsin2p,this implies sin2p=0 and cosp=1. cosp=1 deduces that p=2kπ where kZ. However sin2p=0 implies 2p=lπ where lZ. Hence 2=lπ2kπ=l2kQ,which is impossible. Therefore we get the conclusion.


10. Solution: Additive identity: by definition 0U1 and 0U2, hence 0U1U2.

Closed under addition: if xU1U2 and yU1U2, then xU1 and yU1, hence x+yU1 for U1 is closed under addition. Similarly, x+yU2. Therefore x+yU1U2.

Closed under scalar multiplication: if xU1U2, then xU1. Then for any λF, we have λxU1 since U1 is closed under scalar multiplication. Similarly, λxU2. Therefore λxU1U2.


11. Solution: Assume Ui are subspaces of V, where iI. Now we will show iIUi is a subspace of V.

Additive identity: by definition 0Ui for every iI, hence 0iIUi.

Closed under addition: if xiIUi and yiIUi, then for any given iI, we have xUi and yUi, hence x+yUi for Ui is closed under addition. Therefore x+yiIUi.

Closed under scalar multiplication: if xiIUi, then xUi for any given iI. Then for any λF, we have λxUi since Ui is closed under scalar multiplication. Therefore λxiIUi.


12. Solution: Suppose U and W are two subspaces of V. We argue it by contradiction. If UW is a subspace of V, moreover UW and WU. Consider uUW and wWU, then u+wUW since UW is a subspace of V. Hence u+wU or W. If u+wU, then w=(u+w)uU. We get a contradiction. If u+wW, then u=(u+w)wW. We also get a contradiction. Hence if UW is a subspace of V, we must have UW or WU.

If UW or WU. Without loss of generality, we can assume UW. Then UW=W is obviously a subspace of V.


13. See If a field F is such that |F|>n1| why is V a vector space over F not equal to the union of proper subspaces of V or A question on vector space over an infinite field.


14. Solution: It is clear that U and W are subspaces of F4. Now assume that (x1,x1,y1,y1)U and (x2,x2,x2,y2)W, then (x1,x1,y1,y1)+(x2,x2,x2,y2)=(x1+x2,x1+x2,y1+x2,y1+y2){(x,x,y,z):x,y,zF}. Hence U+W{(x,x,y,z):x,y,zF4}. For any x,y,zF, we have (0,0,yx,yx)U and (x,x,x,z+xy)W. However, (x,x,y,z)=(0,0,yx,yx)+(x,x,x,z+xy)U+W,hence {(x,x,y,z):x,y,zF4}U+W. Combining this with previous argument, it follows that U+W={(x,x,y,z):x,y,zF}.


15. Solution: Because U is a subspace of V, hence closed under addition.Therefore for any x,yU, we have x+yU, i.e. U+UU. Note that if xU, then x=x+0U+U, hence UU+U. Combining with these arguments, it follows that U+U=U.


16. Solution: For xU and yW, because addition in V is commutative, we have x+y=y+xW+U. This implies U+WW+U. Similarly, we have W+UU+W. Hence U+W=W+U.


17. Solution: Note that in V, we have (x+y)+z=x+(y+z). Hence this is similar to Problem 16. Let xiUi, i=1,2,3, then (x1+x2)+x3=x1+(x2+x3)U1+(U2+U3).Since every element in (U1+U2)+U3 can be expressed as the form (x1+x2)+x3, it follows that (U1+U2)+U3U1+(U2+U3). Similarly, we also have U1+(U2+U3)(U1+U2)+U3. Hence (U1+U2)+U3=U1+(U2+U3).


18. Solution: If U is a additive identity, then for any subspace W of V, we have U+W=W. This means UW by the similar arguments in Problem 15 and 16. Hence the only possibility is U=0, in fact this is the case. Hence 0 is the additive identity. Suppose subspace W of V has additive inverses, then there exists a subspace S of V such that W+S=0. This can only happen when W=0 since WW+S.


19. Solution: Here is a counterexample. Let V=U1={(a,b)R2:a,bR}, U2={(a,0)R2:aR} and W={(0,b)R2:bR}. Then it is easy to see that U1+W=U2+W, but U1U2.

See comments for the example from Neven Sajko.


20. Solution: Take W={(0,z,0,w)F4:z,wF}. For any (x,y,z,w)F4, we have(x,y,z,w)=(x,x,z,z)+(0,yx,0,wz)U+Wsince (x,x,z,z)U and (0,yx,0,wz)W. We have F4=U+W.

Moreover, if (x,y,z,w)UW, then we must have x=y and z=w since (x,y,z,w)U.

Similarly, since (x,y,z,w)W, we have x=0 and z=0. Therefore, x=y=0 and z=w=0, hence (x,y,z,w)=(0,0,0,0). It follows that UW={0}. Hence F4=UW by 1.45.


21. Solution: Since we can only choose the first two coordinates arbitrarily in U, we can take W in which the first two coordinates are zero and the last three are variables.

Let W={(0,0,z,w,s)F5:x,yF}. If (x,y,z,w,s)UW, then x=y=0 since (x,y,z,w,s)W. Moreover, we have z=x+y,w=xy,s=2x since (x,y,z,w,s)U. Therefore z=w=s=0, hence UW={0}.

Again, for any (x,y,z,w,s)F5, we have(x,y,z,w,s)=(x,y,x+y,xy,2x)+(0,0,zxy,wx+y,s2x)U+Wsince (x,y,x+y,xy,2x)U and (0,0,zxy,wx+y,s2x)W. Therefore F5=U+W.

Since UW=0, it follows from 1.45 that F5=UW.


22. Solution: Let W1={(0,0,z,0,0)F5:x,yF}, W2={(0,0,0,z,0)F5:x,yF} and W3={(0,0,0,0,z)F5:x,yF}. By the same argument as in Problem 21, we haveF5=UW1W2W3.


23. Solution: Here is a counter example. Let V=R2, U1={(x,0)R2:xR}, U2={(0,y)R2:yR} and W={(z,z)R2:zR}. By the same argument as in Problem 21, one hasV=U1WandV=U2W,however U1U2.

This is exactly the example from Neven Sajko.


24. Solution: Given any fRR, define fe(x) to be f(x)+f(x)2 and define fo(x) to be f(x)f(x)2 for all xR . Then fe,foRR. Moreover, for all xR, we havefe(x)=f(x)+f(x)2=f(x)+f(x)2=fe(x)andfo(x)=f(x)f(x)2=f(x)f(x)2=fo(x).Hence feUe and foUo. Note that we also havef(x)=f(x)+f(x)2+f(x)f(x)2=fe(x)+fo(x),hence f=fe+foUe+Uo. Since we can choose f arbitrarily, one has RR=Ue+Uo.

By 1.45, to show RR=UeUo, it suffices to prove that UeUo={0}. Let fUeUo, then f(x)=f(x) since fUe and f(x)=f(x) since fUo for all xR . Sum up f(x)=f(x) and f(x)=f(x), we have f(x)=0 for all xR. Hence f=0, which implies UeUo={0}.


Linearity

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