# Abstract Nonsense

## Halmos Sections 2,3, and 4

1.

Problem:

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) $y+(x-y)=x$

Proof:

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.

2.

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 $p^n$

3.

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.

4.

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) $\xi_1+\xi_2=1$

Proof:

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 $(1,0,0)+(0,1,0)=(1,1,0)\notin\mathcal{V}$

5.

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?

Proof:

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)