We draw the complex number system as a 2- dimensional plane. But are any of the numbers there mutually orthogonal? Is this pictture misleading?
ortogonality can only come into play when you have an internal product, which the product in the complex plane is not. Take for example $x = i$ then $||x|| = \sqrt{i^2} = i$. So in that regard, no. But there is a notion of norm on the complex plane given by $||x|| = \sqrt{(\re x)^2 + (\re y)^2}$ which is derived from the internal product structure of $\mathbb R^2$ so that gives you a well defined notion of orthogonalyty
Steve, are you asking whether the standard "dot" product of two complex numbers is definable purely in terms of the field structure?
Thank you João and thank you Paul.
My question is really tongue in cheek. My understanding is that the complex numbers form an infinite field. And therefore they do not constitute a vector space. Yet we draw them graphically, looking like a vector space. And we draw axes at right angles. And if I am correct, when using this plane to represent spacetime, we perform rotations on this plane. So there is a lot about this plane that suggests every number has other numbers that are orthogonal. But are there really any numbers a and b, such that ab + ba = 0 ? I am almost happy to say NO! But that would be very bold of me. I am just wondering if we should point out to physics students that this is not a vector space -- don't think of it as one.
By the way João, how do I get your latex to render as good looking maths on these webpages?
A complex number is a number that can be expressed in the form $a + bi$, where $a$ and $b$ are real numbers and i is the imaginary unit, that satisfies the equation $z^2 = -1$, that is, $i^2 = -1$. In this expression, $a$ is the real part and $b$ is the imaginary part of the complex number.
The complex number $a + bi$ can be identified with the point $(a,b)$ in the complex plane.
If $z=x+iy$ and $w=u+iv$ we define the euclidean (real) scalar product as $ \langle z w\rangle_R= xu+ yv$ and complex scalar product as $ \langle z w\rangle_C=z \overline{w}$.
If we set $C=\{ a + bi : a,b \in R \} $, and consider the euclidean (real) scalar product as $ \langle z w\rangle_R= xu+ yv$, then we can identify $C$ by euclidean plane
$R^2$. So it gives you a well defined notion of orthogonalyty.
Nowadays number systems are understood on an algebraic basis. However, when speaking of orthogonality of complex numbers a geometric meaning can be attached to that word. Possibly the book written by J. L. Coolidge in 1924: “The geometry of the complex domain” (https://archive.org/details/geometryofcomple00cool) is known to somebody.
Okay, here's a thought: What I think Steve is asking is whether the field of complex numbers, as an abstract field, has a definable notion of orthogonality. The alternative is that when we represent C in the standard geometric way, we are assuming tacitly that there is structure that is properly richer than the field structure alone. Well, if the dot product is definable, then so is the modulus (norm). But now consider the following example: Pick a nonstandard extension *C of C, say it's an ultrapower using a free ultrafilter over the integers. Then *C and C have the same cardinality, are both algebraically closed fields of characteristic zero, and are hence isomorphic as fields (by classic categoricity results). If || x|| were definable in C, then C and *C, as "normed" fields, would still be isomorphic. But *C has infinitesimals in the sense of *||x||, and C does not, in the sense of ||x||. This gives us a contradiction. ( As a corollary, we now know that the complex conjugate operation on C is also undefinable from the field structure.)
- I use latex notation.
- If for complex number z and w we consider the complex scalar product z\cdot w=z \overline{w}, then z\cdot w=0 if and only if z=0 or w=0. So there is no notion of orthogonalyty.
Something I left out in my argument above: C and *C have the same uncountable cardinality. That's where categoricity comes in. Two countable algebraically closed fields of characteristic 0 can easily fail to be isomorphic.
The complex number system is a beautiful thing to look at. It is rich in structures. With respect to the "usual" addition and multiplication of complex numbers, it is a field.This roughly means that we can freely add and multiply complex numbers. Furthermore free division of nonzero numbers, is permissible with respect to this multiplication. Since, every field is a vector space over it self the set of complex numbers is a vector space over it self. But this is also a vector space over the field of real numbers, which is isomorphic(mirror image of) to the real plane(which we some times call the two dimensional plane). The concept of orthogonality that we discuss for complex numbers is in this" standard" plane, not in the former vector space. In fact, we seldom use the former vector space in our day to day discussion.
Some thing away from our discussion but interesting to add is the following.
The famous "Gelfand-Mazur Theorem" says that any Banach algebra which is also a field must be isomorphic(mirror image of) to the complex vector space.
I think it's fair to say that, yes, there is orthogonality in the complex plane. The usual vector space structure identifies the plane as a two-dimensional vector space over the real numbers, where the components of the vector are the real and imaginary parts of the complex number. So the complex number 3-4i corresponds to the vector (3,-4). If (u,v) and (x,y) are two vectors then their inner product is ux + vy . So, for instance, the complex numbers 1 (the vector (1,0) ) and i (the vector (0,1) ) are orthogonal. The interplay between the vector space structure and the algebraic structure of the complex plane is what makes it so useful in geometry; for instance, it's fairly straightforward (if a little tedious) to prove the nine-point circle theorem using complex numbers.
Dear Moore you think complex plane as real 2 dimensional vector space ok 1 and i are orthogonal as real vectors (1, 0) and (0, 1)
A vector space over a field F is a set V together with two
operations that satisfy the eight axioms. Elements of V are
commonly called vectors. Elements of F are commonly called
scalars. The first operation, called vector addition or simply
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, called scalar
multiplication takes any scalar a and any vector v and gives
another vector av. In this article, the field of scalars denoted
F is either the field of real numbers R or the field of complex
numbers C.
If $F=C$ (res $F=R$ ) we say complex (real) vector space.
Formally, an inner product space is a vector space V over the
field F together with an inner product.
When F = R, conjugate symmetry reduces to symmetry. That is,
\langle x,y \rangle = \langle y,x \rangle for F = R; while for F =
C, \langle x,y \rangle is equal to the complex conjugate.
Notice that conjugate symmetry implies that \langle x,x \rangle is
real for all x, since we have:
\langle x,x \rangle = \overline{\langle x,x \rangle}.
If we have complex vector space (W,C) then (W,R) real
vector space. IF (W,C) is complex inner product space then we
can define (\langle x,y \rangle)_R= Re [(\langle x,y \rangle)_C]
We can consider complex plane C as complex or real vector space.
But since we have multiplication in $C$, it seems natural to
consider C as complex vector space and then we use complex
inner product and we do not have the concept of orthogonality.
There is a saying. I quote. "Experience teaches us not to assume that the obvious is clearly understood." It is better for furtherance of knowledge that we muse again and again over the so called obvious.
In the context of our discussion I wish to add the following. In a vector space if there is also a well defined multiplication among its elements which obeys the standard rules connecting addition and multiplication, say distributive etc. then we call it an algebra. For example the set of ( complex valued) continuous functions is not only a vector space over the field of complex numbers, but it is an algebra too(with respect to point wise multiplication). But not all vector spaces enjoy such property. The set of complex numbers treated as a vector space over the field of real numbers is not an algebra, where as it is an algebra treated as a vector space over it self. So "multiplication on the complex plane" is alright as long as a complex number is treated as single entity, without reference to its Cartesian coordinate system.
I have no immediate answer, even partial, on the difference between complex analysis and vector analysis. It is an interesting question to ponder over.
The inner products on the complex numbers are of the form =kz1*z2 (where k is a positive real number and z1* denotes the conjugate of z1). Thus the only way you can get orthogonality is if one of the two complex numbers z1, z2 is 0.
Rene Ardila, University of Iowa, speaks about complex (Hermitian) inner
product.
Nikos Dimitrios Bagis speaks about real (euclidean) scalar
product. So in that sense both answer are right.
"Given a complex number 'a' then any complex number b=L exp( \pm i \pi /2) a, with L real not zero, is orthogonal to a".
Note that exp( \pm i \pi /2)=\pm i and so b=L i a is
othogonal on a.
Hermitian Inner Product
A Hermitian inner product on a complex vector space V is a
complex-valued bilinear form on V which is antilinear in the
second slot, and is positive definite. That is, it satisfies the
following properties, where z^_ denotes the complex conjugate of
z.
1. =+
2. =+
3. =alpha
4. =alpha^_
5. =^_
6. >=0, with equality only if u=0
The basic example is the form h(z,w)=sum z_iw^__i (1)
on C^n, where z=(z_1,...z_n) and w=(w_1,...,w_n). Note that by writing z_k=x_k+iy_k, it is possible to consider C^n~R^(2n), in which case R[h]
is the Euclidean inner product and I[h] is a nondegenerate alternating bilinear form, i.e., a symplectic form.
I recently became aware of this question and I would like to add another physics viewpoint. Paraphrasing the answer by AK Mishra. "The various axioms of a vector space hold for complex number arithmetic so complex numbers can be interpreted as a vector space with a standard basis of e1 = (1,0) and e2 = (o,i)." The problem for physics is that since i does not represent a quantity, it cannot directly represent a physically measurable space. So what physically measurable direction does it represent? I have wrestled with this question and came up with this answer.
https://www.researchgate.net/publication/280499966_The_imaginary_unit_i_as_the_temporal_directional_component_of_the_complex_position_vector
A proof in section 2 has it as the direction of time. Maybe someone sees what I see.
Article The imaginary unit i as the temporal directional component o...
If you think of the complex numbers as a two dimensional real vectorspace, then the inner product which defines the notion of orthogonality is = Re(z \bar w>. Then two numbers are orthogonal iff. z\bar w is purely imaginary, or if we discard the trivial cases z = 0 or w = 0 ,
(z/|z|) = e^{±\pi/2} (w/|w|).
Lancaster and Tismenetsky (1985) provide the answer to this question. I will summarize it in the following:
First, we shall define a unitary (respectively, Euclidean) space. It is a complex (respectively, real) linear space together with an inner product from Ψ×Ψ to C (respectively, R). Second, let us define the standard inner product for the n-dimensional complex space as
(x,y)=\sum_{i=1}^n x_i \bar{y}_i, (1)
where vectors x, y are n-dimensional complex vectors, and \bar{y}_i is the conjugate of y_i.
Eq. (1) is an extension of the familiar inner product for n-dimensional Euclidian space. For example, in the 3-dimensional Euclidean space, the inner product is
(x,y)=\sum_{i=1}^3 x_i y_i , (2)
where vectors x, y are 3-dimensional real vectors. Then, using Eq. (2), we can define the angle θ between x and y as
cos θ=(x,y)/[\sqrt{(x,x)}\sqrt{(y,y)}]. (3)
However, we cannot usually define the notion of angle using Eq. (1) because, in this case, cos θ is complex-valued in unitary spaces. Nevertheless, we can say that nonzero elements x and y from a unitary space are orthogonal if and only if (x,y) defined by Eq. (1) is zero.
References
Lancaster, P. and Tismenetsky, M. (1985). The theory of matrices, Second edition. New York: Academic Press.
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