Null subspace of a single vector

[vix_cse]

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.

 

[doodle]

If k = 2, then x = [1 -1]^T solves e_2^T x = 0.

 

[HallsofIvy]

I was a bit confused by your reference to "the null space of e_k". 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 e_k 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).

 

[Hurkyl]

$e_k$ is a vector of $k$ $1's.

It sounds more like e_k is a 1xk matrix of 1's.

 

[vix_cse]

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 e<sub>k 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 couldn't figure out how it was so easy.

thanks.</sub>

 

[alice20006]

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.