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Complex Numbers in Linear Algebra

  1. Jul 7, 2011 #1
    I'm working my way through Shaum's Outline on linear algebra and in it they define a complex number as an ordered pair of real numbers (a,b). So given a real number a, its complex counterpart would be (a,0). Operations of addition and multiplication of real numbers work under the correspondence:

    (a,0) + (b,0) = (a + b,0)
    (a,0)*(b,0) = (ab,0)

    I can follow that, but I'm confused how they define i. I know i=(-1)1/2. They define it as:

    i2 = ii = (0,1)(0,1) = (-1,0) = -1

    So am I to assume that any complex number written as (0,b) = -b?
     
  2. jcsd
  3. Jul 7, 2011 #2
    The complex number (a,b) would be written a+bi. I think you might be getting confused because they defined i as i2 = -1, so, as you wrote, i = -11/2
     
  4. Jul 7, 2011 #3
    The next step after defining i is writing a complex number z = a + bi. I'm just a little confused as to their definition of i.

    Here is the first definition of complex numbers, verbatim:
    and then they go on to define i in the manner I posted above. After that,

    Ok, so as I typed that I see that (0,1) is simply i.
     
  5. Jul 7, 2011 #4

    HallsofIvy

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    The one thing you did NOT say was how to multiply two such pairs- you only say "Operations of addition and multiplication of real numbers work under the correspondence:

    (a,0) + (b,0) = (a + b,0)
    (a,0)*(b,0) = (ab,0) "

    The definition of the sum and product of two general such pairs is
    (a, b)+ (c, d)= (a+ c, b+ d) ("component wise" addition)
    (a, b)* (c, d)= (ac- bd, ad+ bd) (which is definitely not "component wise")

    Your text should also show that all the usual "rules of arithmetic" ( addition and multiplication are associative and commutative and multiplication distributes over addition. There are additive and multiplicative identies, every pair has an additive inverse, and every pair except (0, 0) has a multplicative inverse).

    From that definition of multplication (0, 1)*(0, 1)= (0*0- 1*1, 0*1+ 1*0)= (-1, 0). Since we are interpreting the pair (-1,0) to be the number "-1", that says that (0,1)*(0,1)= (0, 1)2= -1 and so (0, 1) is i. Given that we can then say that (a, b)= (a, 0)+ (0, b)= a(1, 0)+ b(0, 1)= a+ bi.

    By the way, here we are defining i to be (0, 1), and then showing that i2= -1, not showing that (0, 1)= (-1)1/2. There are technical problems with "defining" i by "i=(-1)1/2". Every complex number, like every real number, has two square roots. Which of the two roots of -1 is "i"? The real numbers form an "ordered field" so we can distinguish between positive and negative square roots of numbers but the field of complex numbers is NOT an ordered field so it is impossible, a priori, to distinguish between the two roots of -1. Defining i to be the pair (0, 1) avoids that problem. (The other root of -1 is, of course, (0, -1)= -(0, 1)= -i. But to be able to call it "-i" we must be able to first distinguish that root from "i" itself.)
     
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