Understanding C as a Vector Space

[Mdhiggenz]

Homework Statement

We are to show that the set C of complex numbers, with scalar multiplication dened
by α(a + bi) = αa + αbi and addition defined by (a + bi) + (c + di) = (a + c) + (b + d)i,
satises the eight axioms of a vector space

I have a few questions about this problem,

What is the term i? is it just a fancy way of saying a2

Can we think of these as vectors, for instance (a+bi)
is the vector X where a is x1 and bi is x2?

Also I was trying to prove the third axiom which states there exist an element 0 in V such that x+0=x for each xεV.

My logic was let (a+bi)= vector X and (c+di)= Vector Y
X+Y=X

X-X+Y=X-X
Y=0
thus X+Y=X

Thanks for the help guys.

Homework Equations

The Attempt at a Solution

 

[Dick]

Homework Statement

We are to show that the set C of complex numbers, with scalar multiplication dened
by α(a + bi) = αa + αbi and addition defined by (a + bi) + (c + di) = (a + c) + (b + d)i,
satises the eight axioms of a vector space

I have a few questions about this problem,

What is the term i? is it just a fancy way of saying a2

Can we think of these as vectors, for instance (a+bi)
is the vector X where a is x1 and bi is x2?

Also I was trying to prove the third axiom which states there exist an element 0 in V such that x+0=x for each xεV.

My logic was let (a+bi)= vector X and (c+di)= Vector Y
X+Y=X

X-X+Y=X-X
Y=0
thus X+Y=X

Thanks for the help guys.

Homework Equations

The Attempt at a Solution

i is the imaginary unit. i^2=(-1). But all you really need to know about complex numbers is that a+bi=c+di if and only if a=c and b=d. And, sure, the additive identity is 0 or 0+0i. Same thing.

 

[Mdhiggenz]

Thanks for the response Dick. So my logic and reasoning was perfectly find for this problem?

 

[Dick]

Thanks for the response Dick. So my logic and reasoning was perfectly find for this problem?

You deduced the identity is 0+0i, sure. Showing it is an identity is just a matter of saying (0+0i)+(a+bi)=(0+a)+(0+b)i=a+bi. That's only one axiom though. Haven't you got seven more to go?

 

[Mdhiggenz]

Yea, but I just wanted to make sure I was on the right track. Thanks!