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.