My Math Forum Scalar Multiplication in a complex vector Space

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 December 9th, 2009, 08:46 AM #1 Newbie   Joined: Dec 2009 Posts: 2 Thanks: 0 Scalar Multiplication in a complex vector Space Hi all, I am trying to figure out the effect of multiplying a complex vector in C^2 space with a complex scalar. The theory says this produces a parallel vector like the one in real vector space case. For example. If I multiply a real vector [a,b] by a real scalar c, it will be [ac,bc], that is a vector parallel to [a,b]. Similarly, if I multiply a complex vector [a+ib,c+id] by a complex scalar (lets say e+if), will the new vector be parallel to [a+ib,c+id]? How is this described geometrically? Thanks.
December 10th, 2009, 03:07 AM   #2
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Re: Scalar Multiplication in a complex vector Space

Describing C^2 geometrically is difficult, because we're not very good at visualizing 4 dimensions, but...

Quote:
 Originally Posted by Teddy The theory says this produces a parallel vector like the one in real vector space case.
I don't know if parallel is really the correct word, since the two vectors will be overlapping. I would say something like "a vector in the same direction"

Quote:
 For example. If I multiply a real vector [a,b] by a real scalar c, it will be [ac,bc], [*snip*]. Similarly, if I multiply a complex vector [a+ib,c+id] by a complex scalar (lets say e+if), will the new vector be parallel to [a+ib,c+id]? How is this described geometrically?
Yes and no; just like in the real number case, the new vector will be the original times a scalar. Namely, (e+if)[a+bi,c+di] = [(e+if)(a+bi),(e+if)(c+di)]= [ae-bf+(af+be)i, ce-df+(cf+de)i]. So in this sense, it's just the original vector extended/contracted by (e+fi). So it's "parallel" to the original with respect to C.
On the other hand, extending a vector by a non-real "distance" will move the vector around. Think about what happens in C: multiplying by i will rotate the vector by ?/2. In the case of C^2, you sort of have two of these 2-d spaces perpendicular to each other in 4 space, and multiplying by a non-real scalar will rotate the vector in both of these, but they will rotate together.

I guess what I'm trying to say is the geometric intuition sort of breaks down.

December 11th, 2009, 02:48 AM   #3
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Re: Scalar Multiplication in a complex vector Space

Thanks so much cknapp,

But one question:
Quote:
 multiplying by a non-real scalar will rotate the vector in both of these, but they will rotate together.
Thus, if they rotate together, it looks to me that their resultant (sum) rotates together too.
So, why do we say they finally produce a vector in the same direction ?

 December 11th, 2009, 03:08 PM #4 Senior Member   Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3 Re: Scalar Multiplication in a complex vector Space Actually, I'm pretty hazy with the linear algebra, and am probably wrong, but... With respect to C, there is no rotation. But the way we visualize multiplication in C is by rotating in R^2, so we "see" a rotation when trying to visualize it. On the other hand, multiplying by a non-real "length" doesn't seem to make much sense, and my linear algebra isn't good enough to figure out whether it does make sense in some way or not. Hope that helps, and isn't wrong.

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