I found interesting message on the TI discussion list. One of the

users over there complained about the bug in the dot product function

when applied to the vector in complex space. The example given was:

Dot([1+i],[1+i])

On the TI89 it produces 2

On the HP49 it prodcues 2i

Initially many (me including) involved in the discussion thought, that

TI has a bug and HP is right. That was reinforced by the outcome of

the Derive and Mathematica which supposedly produce identical to HP

outcome.

What is interesting, later in the discussion somebody pointed out,

that the dot product when applied to complex spaces requires second

vector to be conjugated. I've checked the web, and in fact in many

respectable math web sites including Wolfram Research this is how the

dot product is "defined" for the complex spaces. That obvioulsy makes

TI answer correct and everybody else (including expensive desktop

application) buggy ! What you guys think about it ? Are there

inconsistent interpretations of the dot product in the math world when

it comes to the complex spaces ? Is traditional definition for the

real space also viable under certain assumptions as a simply another

operator on the complex space that just happen to be named "dot" ?

JM

1. how do i make the dot space dot space that are used in bulletins?

2. Dot Product on Complex Numbers

I found a curious thing while computing the dot product of two

complex numbers.

Suppose I have complex numbers:

a = 2 + 3i

b = 3 + 6i

If I manually compute this, I get: 24 - 3i

However, using dot( a, b ) returns : 24 + 3i

I've tried many different combinations, and in each case, the result

returned from dot is always the complex conjugate of the correct

answer.

According to my linear algebra texts, dot product on complex numbers

is computed by:

(a,b) = a1 * conj ( b1 ) + a2 * conj ( b2 )...

however, it appears matlab is using:

(a,b) = conj ( a1 ) * b1 + conj ( a2 ) * b2 ...

Was this the intended behaviour? Is there a reason for this? Or is

this a bug?

Thanks,

Lesley.

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