Null subspace of a single vector

  • Thread starter vix_cse
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  • #1
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Hi:

was wondering if somebody can help me with this I came across in a paper.

$e_k$ is a vector of $k$ $1's. M is a matrix of size n \times k. The author talks about projecting $M$ onto the null space of $e_k$. This is what confuses me. Which $x$ apart from the 0-vector solves e_kx=0.

any help will be appreciated.
 

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  • #2
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If k = 2, then x = [1 -1]^T solves e_2^T x = 0.
 
  • #3
HallsofIvy
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I was a bit confused by your reference to "the null space of ek". A vector does not have a "null space", a matrix (or linear transformation) has a "null space". But then, reading doodle's response, it became clear that what you really mean was "orthogonal complement": the set of all vectors whose dot product with ek is 0. The orthogonal complement of a vector in n dimensional space is an n-1 dimensional space.

It might help to think of the situation in 3 dimensions, with an x,y,z, Cartesian coordinate system. The vector e1, the unit vector pointing along the x-axis, has orthogonal complement equal to the yz-plane, the set of all vectors of the form (0,y,z).
 
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  • #4
Hurkyl
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$e_k$ is a vector of $k$ $1's.
It sounds more like e_k is a 1xk matrix of 1's.
 
  • #5
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thanks all for your replies. yeah, it does make sense if i think of it as hallsofivy pointed out : the set of all vectors whose dot product with ek is 0.

this is how the author gets to the part in the paper

Z is a n-by-n projection matrix satisfying Z^2 = Z. e is the same as i mentioned.

define Z1 = Z - (1/n) e*transpose(e)

It is easy to see that Z1 represents the projection of the matrix Z onto the null subspace of e.

I couldnt figure out how it was so easy.

thanks.
 
  • #6
I am still confused. As hallsofivy mentioned, the null subspace of e is a set of vectors. By stating that "Z1 represents the projection of the matrix Z onto the null subspace of e.", Z1 should be vectors. But Z1 is matrix. Can anyone explain this? Thanks.
 

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