Abstract Nonsense

Crushing one theorem at a time

Halmos Sections 2,3, and 4



Prove that if \mathcal{V} is a vector space over the field \mathfrak{F}, then for any x,y\in\mathcal{V} and \alpha\in\mathfrak{F} the following are true:

a) 0+x=x

b) -0=0

c) \alpha0=0

d) 0x=0

e) If \alpha x=0 the either \alpha=0 or x=0

f) -x=(-1)x

g) y+(x-y)=x


a) This follows from x+0=x and the commutativity of +

b) We merely note that 0+0=0 and so 0=-0

c) We merely note that \alpha0=\alpha(0+0)=\alpha0+\alpha0 and thus by cancellation \alpha0=0

d) We see that 0x=(0+0)x=0x+0x\implies 0x=0

e) This is identical to the similar problem in the last post.


Problem: If p is a prime then \mathbb{Z}_p^n is a vector space over \mathbb{Z}_p. How many vectors are there in this vector space?

Proof: This is equivalent to asking how many functions are there from \{1,\cdots,n\} to \{1,\cdots,p\} which is p^n


Problem: Let \mathcal{V} be the set of all ordered pairs of real numbers. If x=(\xi_1,\xi_2) and y=(\eta_1,\eta_2) are elements of \mathcal{V} write x+y=(\xi_1+\eta_1,\xi_2+\eta_2), \alpha x=(\alpha\xi_1,0), 0=(0,0) and -x=(-\xi_1,-\xi_2)

Proof: It is not. Notice that (0,1)\ne0 and 1\ne 0 yet 1(0,1)=(1\cdot0,0)=(0,0)=0 which contradicts the e) in the problem one.


Problem: Sometimes a subset of a vector space is itself a vector space. Consider, for example, the vector space \mathbb{C}^3 and the subsets \mathcal{V} of \mathbb{C}^3 consisting of those vectors (\xi_1,\xi_2,\xi_3) such that

a) \xi_1 is real

b) \xi_1=0

c) Either \xi_1=0 or \xi_2=0

d) \xi_1=-\xi_2

e) \xi_1+\xi_2=1


a) This clearly isn’t (remembering that we’re considering \mathbb{C}^3 as being a vector space over \mathbb{C}) since (1,0,0)\in\mathcal{V} but i(1,0,0)=(i,0,0)\notin\mathcal{V}

b) It suffices to show that (0,0,0)\in\mathcal{V}, x,y\in\mathcal{V}\implies x+y\in\mathcal{V}, and \alpha\in\mathbb{C},x\in\mathcal{V}\implies \alpha x\in\mathcal{V} since all the attributes of a vector space (concerning the addition and scalar multiplication) are inherited. But, all three are glaringly obvious. So yes, this is a subspace.

c) No, note that (1,0,0),(0,1,0)\in\mathcal{V} but (1,0,0)+(0,1,0)=(1,1,0)\notin\mathcal{V}

d) Clearly 0\in\mathcal{V}. Also, if x\in\mathcal{V} we have that \alpha x\in\mathcal{V} since \alpha\xi_1+\alpha\xi_2=\alpha(\xi_1+\xi_2)=0. Lastly, if x=(\xi_1,\xi_2,\xi_3),y=(\eta_1,\eta_2,\eta_3)\in\mathcal{V} we see that x+y\in\mathcal{V} since (\xi_1+\eta_1)+(\xi_2+\eta_2)=(\xi_1+\xi_2)+(\eta_1+\eta_2)=0+0=0.

e) No, consider that (1,0,0),(0,1,0)\in\mathcal{V} but (1,0,0)+(0,1,0)=(1,1,0)\notin\mathcal{V}


Problem: Consider the vector space \mathcal{P} (the set of all complex coefficiented polynomials) and the subsets \mathcal{V} consisting those vectors p(x) for which

a) \deg(p(x))=3

b) 2p(0)=p(1)

c) p(x)\geqslant0,\text{  }0\leqslant x\leqslant1

d) p(x)=p(1-x),\quad\forall x\in\mathbb{C}

Which of them are vector spaces?


a) This is not since the zero function isn’t in it.

b) This is.

c) This isn’t since x\in\mathcal{V} but -x\notin\mathcal{V}

d) This is. (maybe, I got a little lazy)


September 22, 2010 - Posted by | Fun Problems, Halmos, Munkres, Topology, Uncategorized | , , , , ,

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