User:J D Bowen/Math725 HW1

1. Let $a, b\in \mathbb{R} \ $. We endeavor to find the multiplicative inverse of the complex number $z=a+bi\in\mathbb{C} \ $. To do this, we assume that at least one of $a, b \neq 0 \ $; otherwise we have $z=0 \ $ and $0^{-1} \ $ is undefined.

We know matrix inverses are easy to compute, and we recall the matrix formulation of the complex numbers as

$a+bi \cong \begin{pmatrix} a & -b \\ b & a \\ \end{pmatrix} \ $.

This matrix has inverse

$\frac{1}{a^2+b^2} \begin{pmatrix} a & b \\ -b & a \\ \end{pmatrix} = \begin{pmatrix} \frac{a}{a^2+b^2} & \frac{b}{a^2+b^2} \\ \frac{-b}{a^2+b^2} & \frac{a}{a^2+b^2} \\ \end{pmatrix} \ $,

which implies that the complex number defined as $c+di, c= \frac{a}{a^2+b^2}, d= \frac{-b}{a^2+b^2} \ $ is the multiplicative inverse of $z=a+bi \ $.

We check our suspicion:

$(a+bi)(c+di) = (a+bi)(\frac{a}{a^2+b^2}-\frac{bi}{a^2+b^2}) = \frac{a^2-abi+abi+b^2}{a^2+b^2} = \frac{a^2+b^2}{a^2+b^2} = 1 \ $, provided $a^2+b^2 \neq 0 \ $, which is guaranteed by our supposition that at least one of $a,b \neq 0 \ $.

2. Let $\mathbf{V} \ $ be a vector space over a field $\mathbb{F} \ $, with $a\in\mathbb{F}, \vec{v}\in\mathbf{V} \ $, and suppose $a\vec{v}=\vec{0} \ $.

Suppose that both $a \neq 0, \vec{v}\neq\vec{0} \ $. Then because $\mathbb{F} \ $ is a field, there exists $a^{-1} \in \mathbb{F} \ $ s.t. $a^{-1}a = aa^{-1} = 1 \ $. Then we can multiply our equation by this inverse to get

$a^{-1}a\vec{v}=a^{-1}\vec{0} \ $

By the identity of scalar multiplication, we have $a^{-1}a\vec{v} = 1\vec{v} = \vec{v} \ $.

Let us know examine the expression $a^{-1}\vec{0} \ $. If we let $\vec{u} \in \mathbf{V} \ $ be any vector, it will be useful to look at the expression $a^{-1}(\vec{0}+\vec{u}) \ $. We can perform the vector sum first and receive $a^{-1}(\vec{0}+\vec{u}) = a^{-1}\vec{u} \ $, or we can distribute and receive $a^{-1}(\vec{0}+\vec{u}) = a^{-1}\vec{0}+a^{-1}\vec{u} \ $. Equating these two results, we have $a^{-1}\vec{u}=a^{-1}\vec{0}+a^{-1}\vec{u} \ $; and the uniqueness of additive identity then tells us that $a^{-1}\vec{0}=\vec{0} \ $.

Applying these two results to our earlier equation $a^{-1}a\vec{v}=a^{-1}\vec{0} \ $, we have $\vec{v}=\vec{0} \ $. This result contradicts our second assumption, that $\vec{v}\neq\vec{0} \ $, and so we reach the conclusion that at least one of our initial assumptions $a \neq 0, \vec{v}\neq\vec{0} \ $ must be false.

3. We inquire as to whether $P=\left\{{(a,b)^t \in\mathbb{R}^2:a\geq 0, b\geq 0 }\right\} \ $, the closed first quadrant of $\mathbb{R}^2 \ $, is a subspace of $\mathbb{R}^2 \ $. Begin by observing that $\mathbb{R}^2 \ $ is a space over the field $\mathbb{R} \ $. By the definition of subspace, we can take an element $-1 \in \mathbb{R} \ $, multiply it by an element $(1,1) \in P \ $, and, if it is a subspace, we should get another element of $P \ $. We perform this multiplication to get $(-1) (1,1)^t = (-1,-1)^t \ $, which fails to be in $P \ $ because $-1<0 \ $. Therefore, $P \ $ is not a subspace of $\mathbb{R}^2 \ $.

4. We inquire as to whether $Q=\left\{{(a,b)^t \in\mathbb{R}^2:a\geq 0, b\geq 0 }\right\} \cup \left\{{(a,b)^t \in\mathbb{R}^2:a\leq 0, b\leq 0 }\right\} \ $, the closed first and third quadrants of $\mathbb{R}^2 \ $, is a subspace of $\mathbb{R}^2 \ $. By the definition of a subspace, we should be able to take the two elements $(1,0)^t, (0,-1)^t \ $, and add them to get another element of $Q \ $. We perform the operation to get $(1,0)^t + (0,-1)^t = (1,-1)^t \ $. This element fails to be $Q \ $ because one element is positive and the other is negative, and so it falls in neither of the two quadrants which compose $Q \ $. Hence, $Q \ $ fails to be a subspace.

5. We inquire as to whether $\mathbb{Z}^2 \ $ is a subspace of $\mathbb{R}^2 \ $. Begin by noting that $\mathbb{R}^2 \ $ is a vector space over the field $\mathbb{R} \ $. If $\mathbb{Z}^2 \ $ is a subspace, then we ought to be able to multiply the vector $(1,1)^t \ $ by an element $\pi\in\mathbb{R} \ $ and receive an element of $\mathbb{Z}^2 \ $. We perform the operation to receive $\pi (1,1)^t = (\pi,\pi)^t \ $, which is not an element of $\mathbb{Z}^2 \ $ since neither co-ordinate is an integer. Hence, $\mathbb{Z}^2 \ $ is not a subspace of $\mathbb{R}^2 \ $.

If we consider $\mathbb{R}^2 \ $ as a vector space over the field $\mathbb{Z} \ $, then we can confirm all the requirements of a subspace for $\mathbb{Z}^2 \ $:


 * $(0,0)^t \in \mathbb{Z}^2 \ $, since 0 is an integer;
 * $(a,b)^t \in \mathbb{Z}^2 \implies (-a,-b)^t \in \mathbb{Z}^2 \ $, since the negative of every integer is an integer;
 * $\forall c \in \mathbb{Z}, c(a,b)^t = (ca,cb)^t \in \mathbb{Z}^2 \ $, since the product of two integers is an integer.

6. We inquire as to whether the set $S = \left\{{a+bi\in\mathbb{C}:a,b\in\mathbb{R},a=b}\right\} \ $ is a subspace of the vector space $\mathbb{C} \ $, when considered both over the field $\mathbb{C} \ $ and the field $\mathbb{R} \ $.

We examine the real case first, and examine the three subspace requirements:


 * Since $0=0 \ $, we have $0=0+0i \in S \ $;
 * Let $\vec{u}=u+ui, \vec{v}=v+vi \in S \ $. Then $\vec{u}+\vec{v}=u+ui+v+vi=(u+v)+(u+v)i \in S \ $.
 * Let $r\in\mathbb{R} \ $ be any real number. Then $r\vec{u}=r(u+ui)=ru+r(ui)=(ru)+(ru)i\in S \ $, since $r,u \in \mathbb{R} \implies ru \in \mathbb{R} \ $.

Now we consider the complex case, and re-examine the third condition. Since now we may multiply a vector $\vec{u}=u+ui, u\in\mathbb{R} \ $ by any complex number $c\in\mathbb{C} \ $, we no longer have the certainty that $(cu)+(cu)i\in S \ $ because $cu \ $ may not be real. Indeed, if we consider the vector $1+i \in S \ $ and the complex number $1+i \ $, when we perform the multiplication we have $(1+i)(1+i)=0+2i \notin S \ $.

7. Theorem: Let $\mathbf{U}, \mathbf{V} \subseteq \mathbf{W} \ $ be subspaces of the vector space $\mathbf{W} \ $ over a field $\mathbb{F} \ $. Then $\mathbf{U}\cap\mathbf{V} \ $ is the largest subspace contained in $\mathbf{U} \ $ and $\mathbf{V} \ $.

Proof: Let $S \ $ be any set contained in $\mathbf{U} \ $ and $\mathbf{V} \ $. In order to be contained in $\mathbf{U} \ $ and $\mathbf{V} \ $, it must be a subset of both. This implies that it is a subset of $\mathbf{U}\cap\mathbf{V} \ $. That is the definition of intersection. Therefore, the largest candidate for a subspace in both $\mathbf{U} \ $ and $\mathbf{V} \ $ is $\mathbf{U}\cap\mathbf{V} \ $, since it is the largest set contained in both $\mathbf{U} \ $ and $\mathbf{V} \ $.

All that remains is to show that $\mathbf{U}\cap\mathbf{V} \ $ is in fact a subspace of $\mathbf{W} \ $. Let $\vec{x}, \vec{y} \in \mathbf{U}\cap\mathbf{V} \ $ be vectors. We examine the requirements of a subspace:


 * Since $\mathbf{U} \ $ is a subspace, $\vec{0}\in\mathbf{U} \ $. Similarly, since $\mathbf{V} \ $ is a subspace, $\vec{0}\in\mathbf{V} \ $.  Therefore, $\vec{0}\in\mathbf{U}\cap\mathbf{V} \ $.


 * $\vec{x},\vec{y}\in\mathbf{U}\cap\mathbf{V}\implies \vec{x},\vec{y}\in\mathbf{U}\implies \vec{x}+\vec{y}\in\mathbf{U} \ $. Similarly, $\vec{x},\vec{y}\in\mathbf{U}\cap\mathbf{V}\implies \vec{x},\vec{y}\in\mathbf{V}\implies \vec{x}+\vec{y}\in\mathbf{V} \ $.  Therefore, $\vec{x}+\vec{y}\in\mathbf{U}\cap\mathbf{V} \ $.


 * Let $a\in\mathbb{F} \ $ be any field element. Then $\vec{x}\in\mathbf{U}\cap\mathbf{V}\implies \vec{x}\in\mathbf{U}\implies c\vec{x}\in\mathbf{U} \ $.  Similarly, $\vec{x}\in\mathbf{U}\cap\mathbf{V}\implies \vec{x}\in\mathbf{V}\implies c\vec{x}\in\mathbf{V} \ $.  Therefore, $c\vec{x}\in\mathbf{U}\cap\mathbf{V} \ $.

Since $\mathbf{U}\cap\mathbf{V} \ $ meets all the requirements for a subspace, it is a subspace. By our earlier logic, it is the largest subspace contained in these two subspaces.