What is meant by compex dimension? (Abstract algebra)

Ineedhelpimbadatphys
Messages
9
Reaction score
2
Homework Statement
Show that the set of n:th order complex polynomials
Pn ≡{a0 +a1z+a2z2 +···+anzn|a0,a1,...,an ∈Cn}
is a vector space. What is its (complex) dimension?
Relevant Equations
Pn ≡{a0 +a1z+a2z2 +···+anzn|a0,a1,...,an ∈Cn}
picture since the text is a little hard to read
IMG_1444.jpeg

i have no problem showing this is a vector space, but what is meant by complex dimention?
Is it just the number on independant complex numbers, so n?
 
Physics news on Phys.org
Do you know how the dimension of a vector space is defined? It involves linear (in-)dependency and therefore the underlying scalar field. E.g. ##\mathbb{C}\cdot [\vec{1}]## is a complex line, hence one-dimensional. But if we consider ##[\vec{1}]## as a real vector, what is the dimension of ##\mathbb{R}\cdot [\vec{1}]##?
 
There are two independant scalars in a complex number. So does that mean 2n.
 
fresh_42 said:
Do you know how the dimension of a vector space is defined? It involves linear (in-)dependency and therefore the underlying scalar field. E.g. ##\mathbb{C}\cdot [\vec{1}]## is a complex line, hence one-dimensional. But if we consider ##[\vec{1}]## as a real vector, what is the dimension of ##\mathbb{R}\cdot [\vec{1}]##?
Ineedhelpimbadatphys said:
There are two independant scalars in a complex number. So does that mean 2n.
Sorry, this was supposed to be a reply. Im really not understanding the subject so sorry for simple questions.
 
Ineedhelpimbadatphys said:
There are two independant scalars in a complex number. So does that mean 2n.
No. It only means ##2n## over the reals! It is still ##n## over the complex numbers.

If we write a complex number ##a+\mathrm{i} b## as real vector ##(a,b)## then we get
$$
\dim_\mathbb{R} \left\{\mathbb{R}\cdot\begin{pmatrix}a\\0\end{pmatrix}\oplus \mathbb{R}\cdot\begin{pmatrix}0\\b\end{pmatrix}\right\}=2\, , \,\dim_\mathbb{R} \mathbb{R}\cdot\begin{pmatrix}a\\b\end{pmatrix}=1
$$
and
$$
\dim_\mathbb{C} \left\{\mathbb{C}\cdot a + \mathbb{C}\cdot \mathrm{i}b\right\}=1\, , \,\dim_\mathbb{C} \mathbb{C}\cdot (a+\mathrm{i}b) =1
$$
I assume the exercise was to understand this difference. The field is essential here.
 
Last edited:
  • Like
Likes FactChecker
The dimension is ##n+1##.
 
Last edited by a moderator:
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top