Complex numbers are essentially 2-dimensional vectors with product defined such that the set is a field and extends real product. So, is there an analogous definition for 3, or higher, dimensional vectors such that the Euclidean space becomes a field also?
123553
If complex numbers are in some sense 2 dimensional numbers, would it be useful or logical to extend the complex number system to a system of 3-dimensional or even n-dimensional numbers? If this does not make sense in general, why does it make sense only to extend from the 1st dimension to the 2nd dimension? What is special about 2-dimensional numbers that they are needed but n-dimensional numbers are not? I know that we have $R^k$, which is in some sense a field of $k$ dimensional numbers, but that is very different from the way in which complex numbers are constructed. So my question is could the complex numbers be extended to a 3rd dimension (or even an nth dimension) in the same way that the real numbers are extended to the complex numbers, in which the properties of complex numbers are preserved as a subset of this higher dimensional space?
108374
Consider the complex vector space $\mathbb{C}^3$ and the subset $$S = \{(1, 0, i), (i, 2, -3), (2-i, 1+i, i)\}$$  Let $\textsf V$ denote $\operatorname{span}(S)$. Find a subset of $S$ that forms a basis for $\textsf V$.  How do I find the subset for something? I can put the vectors into matrix form and make them equal to $x, y, z$ respectfully but from there, I don't know how to proceed.  Am I missing a definition and and a series of steps? I know that a basis in simple terms is :     The minimal amount of vectors that are linearly independent and spans    $\textsf V$.
15850
Quick question: Can I define some inner product on any arbitrary vector space such that it becomes an inner product space? If yes, how can I prove this? If no, what would be a counter example? Thanks a lot in advance.
83432
Explain why $\mathbb{Z \times Z}$ and $\mathbb{R \times R}$ is not a field and that why any external direct sum of two fields cannot be a field. I believe it has much to do with the lack of every non-zero element having an inverse, however I am having difficulty seeing it.
6057
As normed division algebras, when we go from the complex numbers to the quaternions, we lose commutativity. Moving on to the octonions, we lose associativity. Is there some analogous property that we lose moving from the reals to the complex numbers?
19864
In my opinion both are almost same. However there should be some differenes like any two elements can be multiplied in a field but it is not allowed in vector space as only scalar multiplication is allowed where scalars are from the field.  Could anyone give me atleast one counter- example where field and vector space are both same. Every field is a vector space but not every vectorspace is a field. I need an example for which a vector space is also a field.  Thanks in advance. (I'm not from mathematical background.)
76918
Complex numbers are usually formally defined as pairs of real numbers. Although there are operations on $\mathbb{C}$, such as complex multiplication, which are not found in operations usually applied to $\mathbb{R}^2$, the sets themselves seem to be the same. Each consists of pairs of real numbers.  So is it okay to say that $\mathbb{C} = \mathbb{R}^2$? It seems formally correct, but something doesn't feel quite right about it.
2257
Suppose that we are dealing with the real field $\mathbb{R}$ and we define $T:\mathbb R^2\to\mathbb R^2$ by $$T(u,v):=(-v,u).$$ It is trivial to see that $\langle(u,v),T(u,v)\rangle = 0$ for all $u, v \in \mathbb{R}$, where $\langle \, \cdot \, , \, \cdot \, \rangle$ is the standard inner product.  A question in Hoffmann-Kunze asks us to define all possible inner products $[ \, \cdot \, , \, \cdot \, ]$ such that $[\alpha, T(\alpha)]=0$ for all $\alpha\in\mathbb R^2$ and $T$ is as above. How does one find all such inner products?
76686
Today I proved that if $V$ is a normed space with norm $\|\cdot\|$ then I can define a norm on $V \times V$ that induces the same topology as the product topology as follows: $\| (v,w) \|_{V \times V} = \|v\| + \|w\|$.  I think I can do the same for an infinite product $V^{\mathbb N}$ by defining $\|(v_n)\|_{\mathbb N} = \sum_{n=0}^\infty \frac{1}{2^n} \|v_n\|$ and I proved it using the proof of the case $V \times V$ and changing some minor things.  Can you confirm that this result is correct? Thanks.
99750
Let $\mathcal F$ be a field. Suppose that there is a set $P \subset \mathcal F$ which satisfies the following properties:   For each $x \in \mathcal F$, exactly one of the following statements holds: $x \in P$, $-x \in P$, $x =0$.  For $x,y \in P$, $xy \in P$ and $x+y \in P$.    If such a $P$ exists, then $\mathcal F$ is an ordered field.   Define $x \le y \Leftrightarrow y -x \in P \vee x = y$.   Exercise: Prove that the field of complex numbers $\mathbb C$ cannot be given the structure of an ordered field.   My Work So Far: (Edit 1 note: This section and the Question is at the beginning, simply leaving this up for reference as to where I started)  Let $i$ be such that $i \in P, i \ne 0 \Rightarrow i &gt; 0$. But $i^2 = -1 \notin P$.   My Question: I am not sure how much I need to redefine, and how I go about rigorously making this patchwork argument airtight. I am aware that I have not addressed how I assumed that $-1 \notin P$, but I'm not sure how to distinguish between $1$ and $i$ in this proof.     Edit #1    1st Step: Showing that $-1 \notin P$, observe that $(-1)(-1) = 1$ therefore if $-1 \in P$, both $x, -x \in P$, a contradiction.   2nd Step: To show $i \notin P$, we have that if $i \in P \Rightarrow i^2 \in P$, but $i^2 = -1 \notin P$, so $i \notin P$.   3rd Step: To show $-i \notin P$, we have $(-i)(-i) = i^2 \notin P$, so $-i$ cannot be in $P$.   Conclusion: Since $i \ne 0$, and $i, -i \notin P$, there is no set $P \subset \mathbb C$ that satisfies the above properties, thus $\mathbb C$ is not ordered.   Thank you André Nicolas and Eric Stucky for your help!
28943
I read that the cross product can't be generalized to $\mathbb R^n$. Then I found that in $n=7$ there is a Cross product:   Why is it not possible to define a cross product for other dimensions $ \ge 4$?
35285
Is there a canonical way to define on any vector space over $\mathbb{K}=\mathbb{R},\mathbb{C}$ a norm ? (Or, if there isn't, can someone give me an example of a vector space over $\mathbb{K}$ that is not normable ?)  I have now looked through several books on the subject but nowhere is something like this mentioned and I also wasn't able to find a way to construct such norm (or to find a counterexample).
97485
I've tried a method similar to showing that $\mathbb{Q}(\sqrt2, \sqrt3)$ is a primitive field extension, but the cube root of 2 just makes it a nightmare.  Thanks in advance
15835
I've read a bit about the history of the complex numbers, and many seem to credit Caspar Wessel with the idea of associating the complex numbers as points on a 2-dimensional plane (or at least the first to explicitly publish the idea). But what was the motivation behind this? Why did people make this leap, and what were their justifications for this?  edit: just to be clear, my question is only about the justification of identifying complex numbers as points on a plane, NOT about justifying complex numbers themselves.
83713
I was looking at the definition on wikipedia of a vector space (similar/equivalent definitions are everywhere, but I thought I'd list it here for completion):  A vector space over a field $F$ is a set $V$ together with two binary operations that satisfy the eight axioms listed below. Elements of $V$ are called vectors. Elements of $F$ are called scalars. In the two examples above, our set consists of the planar arrows with fixed starting point and of pairs of real numbers, respectively, while our field is the real numbers. The first operation, vector addition, takes any two vectors $v$ and $w$ and assigns to them a third vector which is commonly written as $v + w$, and called the sum of these two vectors. The second operation takes any scalar a and any vector $v$ and gives another vector $av$. In view of the first example, where the multiplication is done by rescaling the vector $v$ by a scalar $a$, the multiplication is called scalar multiplication of $v$ by $a$.  To qualify as a vector space, the set $V$ and the operations of addition and multiplication must adhere to a number of requirements called axioms. In the list below, let $u$, $v$ and $w$ be arbitrary vectors in $V$, and $a$ and $b$ scalars in $F$.   Associativity of addition: $u + (v + w) = (u + v) + w$ Commutativity of addition: $u + v = v + u$ Identity element of addition: There exists an element $0 ∈ V$, called the zero vector, such that $v + 0 = v$ for all $v ∈ V$. Inverse elements of addition: For every $v ∈ V$, there exists an element $−v ∈ V$, called the additive inverse of $v$, such that $v + (−v) = 0$ Compatibility of scalar multiplication with field multiplication: $a(bv) = (ab)v$ Identity element of scalar multiplication: $1v = v$, where $1$ denotes the multiplicative identity in $F$. Distributivity of scalar multiplication with respect to vector addition: $a(u + v) = au + av$ Distributivity of scalar multiplication with respect to field addition: $(a + b)v = av + bv$   I was wondering if there were any examples of vector spaces where only the law for the identity element of scalar multiplication fails? The only one I could think of would be to redefine a true vector space's scalar multiplication to multiply the result by a constant factor. Are there others?
94549
Is there any way to extend the elementary definition of powers to the case of complex numbers?   By "elementary" I am referring to the definition based on $$a^n=\underbrace{a\cdot a\cdots a}_{n\;\text{factors}}.$$ (Meaning I am not interested in the power series or "compound interest" definitions.) This is extended to negative numbers, fractions, and finally irrationals by letting $$a^r=\lim_{n\to\infty} a^{r_n}$$ where $r_n$ is rational and approaches $r$.  For a concrete example, how would we interpret $e^i$ in terms of these ideas?
88273
Can we define a vector space structure on $\mathbb {R}^n$ other than usual scalar multiplication and usual addition such that the dimension of $\mathbb {R}^n$ over $\mathbb {R}$ is not $n$ but some $m$ not equal to $n$?