
Linear Algebra Linear Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
April 30th, 2015, 12:47 AM  #1 
Newbie Joined: Mar 2015 From: Africa Posts: 17 Thanks: 0 Math Focus: Dont' know which area yet, currently enrolled for BSc in Mathematics and Applied Mathmatics  3rd y.  Basis for Complex Vector Space over the Complex and then the Real Fields
I am trying to make sense of the basis for the complex vector space C over the field of scalars F, where F = C (complex field), and then F = R (real field). Intuitively for me I would have assumed B = {1,i} for C over F = C and B = {1} for F = R. This would yield dim(C(C)) = 2 and dim(C(R)) = 1, which, again intuitively makes sense to me. But it is the other way around. I have worked through the proofs but I still don't click it properly, I think. So far I understand it as follows: When F = C, the vector in the vector space C, although complex a + ib, can be viewed as a single 'bag of things'. When F = R, the vector is the vector space C, still complex a + ib, is seen as being made up out of two 'things'. I am still not satisfied though. Can anyone offer some more insights? 
April 30th, 2015, 04:00 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,940 Thanks: 2266 Math Focus: Mainly analysis and algebra 
Suppose we have a basis vector $\newcommand{\i}{\mathrm i}a + b\i$ (where $a$ and $b$ are nonzero real numbers). Then with complex coefficients we can form $$(x + y\i)(a + bi) = (ax  by) + (bx + ay)\i$$where $x$ and $y$ are real numbers. Now, if we wish to make some vector $r + s\i$ ($r$ and $s$ both real) we have$$r = ax  by \qquad s=bx + ay$$ which means that we choose$$x = {arbs \over a^2  b^2} \qquad y = {asbr \over a^2  b^2}$$ The vector $a + b\i$ therefore spans the vector space $\mathbb C$ as required. Since this basis has only one vector, the dimension is 1. With real coefficients we have $$a = b\i = a(1 + {b \over a}\i)$$ (where $a\ne 0$) and any real multiple of this has the imaginary part in fixed proportion ${b \over a}$ to the real part. We cannot therefore span the space $\mathbb C$ with a single vector. We can do a similar analysis for $b \ne 0$. Last edited by v8archie; April 30th, 2015 at 04:08 AM. 

Tags 
basis, complex, fields, real, space, vector 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Basis for Vector Space  MQ1993  Algebra  1  March 11th, 2015 09:20 PM 
Is basis(row space) = basis(vector space)?  PhizKid  Linear Algebra  1  November 25th, 2013 07:56 AM 
Basis of complex vector space  shine123  Linear Algebra  1  October 16th, 2012 06:50 AM 
Scalar Multiplication in a complex vector Space  Teddy  Linear Algebra  3  December 11th, 2009 03:08 PM 
basis to a vector space  OriaG  Calculus  0  December 31st, 1969 04:00 PM 