Complex Vector Space Puns

Before anyone says anything, we unfortunately CAN’T use the Cayley Hamilton theorem (We’re proving a special case of it actually)

So this is a part of a bigger proof on my homework, I have already reliably proven that any operator, S, on a complex vector space has it’s characteristic polynomial=0 evaluated at S

We assume the theorem that says if B is a basis for V, then the vector space of matrices Mn(F) is isomorphic to L(V,V) by taking T to it’s matrix representation in Mn(F) with respect to the B basis.

Can anyone check over a part of my proof for me? I feel as though I’m overlooking something when it comes to viewing polynomials in F[t] in sets like C[t]

excerpt of my proof found here

Any help would be greatly appreciated

So, let's assume a vector space V=R^n over the field K of real numbers, K=R. Now let's define an arbitrary linear operator t: R^n -> R^n with its matrix A, such that t(x) = Ax for any vector x in R^n.

The eigenvalues of a matrix are defined as a scalar value by which our corresponding eigenvector is scaled and since we're working over the field of real numbers, one would expect any eigenvalue to be real. However, as we all know, it's really simple to end up with a complex eigenvalue, even for some simple matrices. How is this possible?

This may seem dumb to y'all, so pls be easy on me. There's probably something obvious I'm missing.

Ok, I just don't understand how every complex vector space is also a real vector space. Isn't a real vector space a set of vectors of which has entries of only real numbers. And, a complex vector space is a set of vectors of which all have entries of complex numbers. So, how could every complex vector space also be a real vector space? Is there something obvious I'm missing?

That is, if V is a finite dimensional real vector space with an operator I: V -> V so that I^2 = -1, what are the bases of the eigenspaces I in the complexification V ⊗ C = V^C ?

The eigenspaces of $I$ are pretty simple. They are elements of the complexification of V of the form v - iIv with eigenvalue i and v + iIv with eigenvalue -i. I'll denote these eigenspaces V^(1,0) and V^(0,1) respectively. Clearly, their intesection is 0. I would like to show that V^C = V^(1,0) ⊕ V^(0,1). The only thing left to do then is to show that V^(1,0) + V^(0,1) spans V. This seems pretty easy though since for v \in V, we have 2v = (v + iIv) + (v - iIv).

I believe this proves V^C = V^(1,0) ⊕ V^(0,1). My point of confusion here is that my construction doesn't give a nice basis for V^(1,0) and V^(0,1). I would expect their dimensions to be equal but without a basis, it's hard to get a good grasp as to why that's the case. To prove that, I would just use the fact(?) that the map v + iIv → v - iIv is a linear isomorphism between V^(1,0) and V^(0,1). This would indeed give us a basis for the other given a basis for one but I'm still at a loss for what might be the most natural basis on V^(1,0). What do you think?

Thank you very much for your help!

I was reading about linear forms, dual basis, bilinear forms and tensor product, but I am concerned about, what does happen with all this items when V is a complex vector space?

How multilinear algebra and tensor calculus work with complex functions and numbers?

This is probably a stupid question, but here we go anyway.

Let's view the complex number C as a vector space over the real number. That is, C = span{1, i}.

Now define an inner product on this space like this: https://imgur.com/PJTdb5B

That is, <a+bi, c+di> = (a+bi)(c-di)

Clearly, this defines an inner product.

Let W = span{1}. What's the Orthogonal complement of W?

From the definition, it seems the Orthogonal complement of W is just {0}, but that wouldn't make sense since C here is 2 dimensional space, so the Orthogonal complement of W must have dimension of 1, since W has dimension of 1.

What am I missing?

Thanks for your time.

I am wondering why when we take an inner product of two complex vectors we need to complex conjugate one of them? How is this rule for inner product on C^n a natural extension from inner products in R^n? Thanks!

Hi reddit, I have hust worked through the proof that every finite dimensional normed vectorspace over the real numbers is equivalent.

I wondered if the same is true for Vectorspaces over the complex numbers, since you can easily proof that the max norm dominates every other norm and I never saw a part in the proof that every norm dominates the max norm, where we use the fact that we are in a Vectorspace over the reals.

The idea of the other proof was that we look at the sphere of all elements where the maxnorm is 1 and get a contradiction if we assume that the other norm gets arbitrarily small on this set. We only use that every sequence in this set has a convergent sequence in it and that both norms are continous on V with the max norm.

Have I missed something? Usually our professor generalizes quite a lot so I think I must have.

I want to start out by saying I am a physicist, so the majority of my math knowledge comes from what I have learned in my physics classes. We do take mathematical methods courses, however obviously they are tailored toward what a physicist needs to know and therefore aren’t always as in depth or rigorous as the same subject taught to a pure mathematician.

My question however is in trying to understand what exactly the difference between a complex vector space and a real vector space is, and is there always an isomorphism between complex vector spaces of n-dimensions and real vector spaces of 2n dimensions? I feel I have a very firm grasp of what exactly constitutes a vector space, the axioms that must be satisfied, and that vector spaces must be closed under vector addition and scalar multiplication. I also understand that every vector space comes along with a field and that field can either be real or complex. My confusion however is that if I consider R^2, the space of 2 dimensional real vectors, I think of this as a set of 2 ordered points with (obviously) dimension 2 and I can write a basis as {(1,0),(0,1)}. If I consider C^1, the set of 1 dimensional complex vectors, this is now a 1 dimensional space for which I can write a basis as {(i)} and which any further vector will be a complex scalar multiple of this basis vector. The thing I am confused about is that I inherently think of C^1 as being the complex plane, which seems to be 2 dimensional, having 2 degrees of freedom, if I can use that physics term there. So why are there 2 degrees of freedom in R^2, but 1 degree of freedom in C^1? Does the extra degree of freedom “transfer” from the set of basis vectors in the case of R^2 to the field of scalars in C^1? Am I looking at this completely wrong and just don’t realize it?

The real reason I am curious is that I have been learning about quaternions which form a basis for spin-space (Pauli matrices), and that a quaternion field can be viewed as C^2 or R^4, so is it always the case you can look at a complex vector n-dimensional space as a real 2n dimensional space?

Also, I apologize if I have misused any terms here. Please feel free to correct me, as I am trying to learn the correct terminology which isn’t always easy whenever its just you reading out of a textbook without any feedback.

> Given any unitary matrix, prove that the modulus squared of each of the entries forms a doubly stochastic matrix.

I computed this using a simple 2x2 matrix with entries [[a + bi, c + di], [e + fi, g + hi]] . I arrived at the following conclusion by adding rows/cols and creating system of equations-

a + bi = g + hi
e + fi = c + di

This tells me that (a, g), (b, h), (e, c), (f, d) need to be equivalent. The answer in the back of the book had the following solutions that make sense. I don't know how to reconcile any of the preceding ordered pairs aren't equal changing the value of the Kronecker delta function. I understand that my 2x2 matrix is not general enough for a rigorous proof but I needed to start somewhere.

(UxU†)[j, k] = sigma(i) { U[ j, i ] x U†[i, k] }
= sigma(i) { U[j, i] x conjugate(U[k, i]) } = δ(j,k)

.

sigma(k) { |U[j, k]|^2 }=
sigma(k) { U[ j, k] × conjugate(U[j, k]) }
= δ(j, j) = 1

Note- this is a selfstudy and it's been ages since I last used linear algebra. This is my first proof in a complex vector space.

I was certain they could until my mathematics book: http://imgur.com/XGcUOso seemed to differ in opinion.

I can follow the definitions of complex vector spaces, but it is harder to get an intuitive understanding of it. Any comments will be appreciated.

I have been using OpenMP reduction with std::vector for a long time. The function below generates an error message: user defined reduction not found for ‘res’. I could not figure out the cause. My compiler is g++ (GCC) 11.2.1. Any suggestions will be greatly appreciated.

#include <vector>
#include <cmath>
#include <algorithm>
#include <complex>
using namespace std;

complex<double> complex_func(vector<complex<double>> x)
{
  long n = x.size();
  complex<double> res = 0.0;

  #pragma omp parallel for reduction(+:res)
  for (long i = 0; i < n; i++)
  {
    res += exp(pow(log(x[i]), 2));
  }
  
  return res;
}

When we say "complex vector" is it the same as a "complex number": (Re, Imag) = a + bi ? More specifically if v is a vector in an abstract (complex) vector space V, do we mean that v is a complex number like I mentioned before, or is it a vector with complex components, ie: (a+bi, c+di, ...) ? The terminology here is confusing me.

I understand that C^n can be thought of as R^2n but I do not understand why using complex scalars instead of real scalars cuts the dimensions of the vector space in half. Why is the basis of a complex vector space c^n across a complex field {(1,0,0...,0),(0,1,0,...0),...,(0,....,1)} as there is no imaginary part to these numbers?

Edit: I think I am getting closer to understanding it. C^2 means two-tuples of the form (a+bi,c+di) so if you have real scalars you need 4 separate vectors as part of you basis to form this and so dimR( C^2 ) = 4. Am I correct in saying the basis would be {a(1,0),b(i,0),c(0,1),d(0,i)|a,b,c,d ϵ R) and that I am allowed to assign any value I so choose to a,b,c,d? But if you had complex scalars the basis would be {(a)(1,0),(b)(0,1)|a,b ϵ C} for this I don't understand how I can assign any value I choose to a and b as if I choose a=1+i and b=2+3i this would not be a spanning set of the vector space C^2 as you would not be able to get to a point such as (1,1) with this basis.

Edit 2: I understand now that the basis of C^2 in a complex field is any set of the form {a(1,0),b(0,1)|a,b ∈ R} as you can always form a linear combination of these two vectors with complex coefficients to obtain any two-tuple of the form (a+bi),(c+di)

I've been reading Linear Algebra by Niels Vigand Petersen, and i noticed that he noted that R^n is not a vector space over the complex numbers if n > 0. I can not see why that is the case, because to me it looks like it fulfills the criteria. For example, given vector [1,2] and vector [ 3,4], and also complex number 2i and 3i, they seem to fulfill the criteria, AKA:

2i( [1,2] + [3,4]) = 2i*[1,2] + 2i*[3,4]

(2i + 3i) * [1,2] = 2i*[1,2] + 3i*[1,2]

(2i3i)[1,2]=2i*(3i*[1,2])

I can't see what i'm missing here.

Probably the math doesn't hold up, but I've never quite grasped the concept of complex numbers and always assumed it was kind of a bucket of 2 distinct values, one that can "mingle" with real numbers and one that cannot (or well that is "0" by default for non C numbers).

Is that view total non-sense or is there a parallel to make?

I am trying to get at the essence of a vector space, but each answer uses examples involving (real) numbers. Or functions of real numbers.

I want to see if we can construct a system, using the concepts of vectors in a vector space, and not have it directly involve numbers at all. Is this possible? Basically, can vectors/vector-spaces be used to model things outside of numbers?

I get that a vector space requires satisfying some 8 axioms, or as Wikipedia says, "a vector space over a field F is a set V together with two operations that satisfy the eight axioms listed below."

Can you construct a system of vectors in a vector space which satisfies these properties/axioms, and yet doesn't directly revolve entirely around numbers (so isn't using the real numbers directly, or functions of real numbers, or matrices of real numbers, etc.). Is it possible?

My starting of an example would be something along the lines of... Say for example we try to make the vectors be molecules. Can it be done? In this way, they will not be directly related to numbers.

• Associativity of vector addition (example, elements are acid/base/water, acid + base = neutral element)
• Commutativity of vector addition (acid + base = base + acid)
• Identity element of vector addition (v + 0 = v, maybe perhaps an unreactive molecule)
• Inverse elements of vector addition (v + (−v) = 0, acid + its opposing base = water)
• Compatibility of scalar multiplication with field multiplication (a(bv) = (ab)v, dunno)
• Identity element of scalar multiplication (would have to think harder how to fit the rest into the mold...)
• Distributivity of scalar multiplication with respect to vector addition ...
• Distributivity of scalar multiplication with respect to field addition ...

Maybe instead of molecules, we use some other objects like light waves or colors, or human beings, etc. Can vector spaces have as their "vectors" arrays of arbitrary non-numerically-related objects like these? If so, what is a complete example?

At all costs, please don't write about something to do with numbers, the real numbers, or anything directly referencing the real numbers.

If it can't be done, why not? If it can be done, what is an example, and if you can, what is a practical application of vectors as NOT-numbers in the real-w

Basically the title. I can't find it and Desmos has a different one. I'm looking specifically to the vector plotting way, i like that the best.

Hello

Currently I'm studying some linear algebra and complex numbers. If I'm correct on this, you can represent complex numbers as vectors in a vector space C over R with the basis vectors (1,0) and (0,i).

I am currently stuck however on multiplication and how to visualize this with vectors. I get how you multiply complex numbers in their number form, but this doesn't seem possible with vectors. The dot product gives you a number (the length of the projection of one vector on the other if I remember correctly) and the cross product is undefined in 2D space.

Could anyone help me with this? Thank you

In Quantum Mechanis, Eigenvalues are degenerate if if they correspond to more than one direction. If we look at the psi function psi(x,t)=Ae^i/hbar(px-Et), what would be an example of a direction? What constitutes as a direction in this space? Can you give an example of both degenerate and non-degenerate Eigenvalues of the psi function? Thanks in advance!

Let V be a vector space over the complex numbers that has a countable basis, and suppose A : V → V is a linear transformation.

Prove that there must exist a complex number λ such that the linear transformation A - λI is not invertible, where I is the identity transformation.

I was reading about linear forms, dual basis, bilinear forms and tensor product, but I am concerned about, what does happen with all this items when V is a complex vector space?

How multilinear algebra and tensor calculus work with complex functions and numbers?