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Projection of a on b when a, b are complex

  1. Nov 14, 2014 #1
    see figure: http://en.wikipedia.org/wiki/Vector_projection#mediaviewer/File:Projection_and_rejection.png

    Im reading about projections of vectors. My book says nothing about what the projection a1 of a on b is when a and b are complex vectors.

    To find the formula for the projection, one needs to take the scalar product of b and (a-a1) and set it to equal zero and set that a1 equals some number times b.

    But since a and b are complex, doesn't it matter which order this is done in?

    If both a and b are complex, do we get two different projections onto b from a ?

    see this, about 9:30 minutes in: http://www.khanacademy.org/math/lin..._trans_examples/v/introduction-to-projections
     
    Last edited: Nov 14, 2014
  2. jcsd
  3. Nov 14, 2014 #2

    jedishrfu

    Staff: Mentor

  4. Nov 14, 2014 #3

    Fredrik

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    Let ##V## be a complex vector space. Let ##\{e_i\}_{i=1}^n## be an orthonormal basis for it. I'll use the physicist's convention for inner products. This means that the inner product is linear in the second variable and antilinear (=conjugate linear) in the first. Let ##v\in V## be arbitrary. Now let's write ##v=\sum_i v_i e_i##, and then compute ##\langle e_i,v\rangle##.
    $$\langle e_i,v\rangle = \left\langle e_i,\sum_j v_j e_j\right\rangle =\sum_j v_j \langle e_i,e_j\rangle =v_i.$$ This means that every ##v\in V## can be written as ##v=\sum_i \langle e_i,v\rangle e_i##. Now we can see that the projection of ##v## onto the 1-dimensional subspace spanned by ##e_i## is ##\langle e_i,v\rangle e_i##. It can't be ##\langle v,e_i\rangle e_i##.

    Edit: I took a quick look at the video and saw that it's talking about the dot product in ##\mathbb R^2##. So you may not be familiar with notations like ##\langle x,y\rangle##. This is the standard notation for the inner product of x and y. An inner product is defined as a function that takes two vectors as input and gives you a number as output. (There's also a list of conditions that it has to satisfy). The dot product on ##\mathbb R^2## is such a function. So the dot product is a special kind of inner product.
     
    Last edited: Nov 14, 2014
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