Re: ? orthogonal hyperplane
- From: "I.N. Galidakis" <morpheus@xxxxxxxxxxxx>
- Date: Mon, 21 Jul 2008 13:06:18 +0300
asecant@xxxxxxxxx wrote:
Hi:
Given a vector, in 2D, there are two vectors being
orthogonal to that given vector and they are lying on
a line. In 3D, we have inf many such vectors being
orthogonal to that given vector and they all lie on an
orthogonal plane.
My question is: how about in a dimension that is higher
than 3D? What will be the dimensions of the object (or
maybe a hyperplane) that contains all the orthogonal
vectors to that given vector?
If you call V the subspace of R^n consisting of all the orthogonal vectors to a
given vector X \in R^n, then V's dimension will be such that dim(R^n/V)=1.
Another way to look at it using the fact that the dot product between X = [x_1,
x_2, ..., x_k] \in V and any vector A = [a_1, a_2, ..., a_k], will be zero:
V in R^2: a_1*x_1 + a_2*x_2 = 0. dim = 1 (line)
V in R^3: a_1*x_1 + a_2*x_2 + a_3*x_3 = 0. dim = 2 (plane)
....
V in R^n: a_i*x_i = 0 (using Einstein summation convention). dim = n-1
(hyperplane)
See:
http://mathworld.wolfram.com/Hyperplane.html
Thanks,--
I.N. Galidakis
.
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