Projection of a on b when a, b are complex

[johann1301]

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

 

[jedishrfu]

Here's a discussion here that may help:

http://en.m.wikipedia.org/wiki/Dot_product

It seems that the dot product is modified to prevent the zero vector case you mentioned which makes it non symmetric.

 

[Fredrik]

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.