Can the null space of matrix B be used for data projection?

Sue_2010
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Hello everyone,

If I have a collection of data points (vectors), and x and y are two vectors among them. I want to project the data to a direction that the Euclidean distance between x and y is Maximally preserved. Then this direction should be the row space of (x-y)’, denoted as row( (x-y)’ ), right? (suppose column vectors.)

Now, suppose I have n pairs of such data points grouped as a matrix A, say

A = [
(x1 – y1)’
(x2 – y2)’

(xn –yn)’
]

For my problem, directly calculate the row space projection matrix P1 = A’*inv(AA’)*A is difficult, so I want to do some approximation.

Suppose all the vectors have been normalized, that is norm(x1) = norm(x2) = … = norm(xn) = norm(y1) = .. = norm(yn). I define matrix B as

B = [
(x1 + y1)’
(x2 + y2)’

(xn + yn)’
]

And plan to find the null space of B, denoted as null(B). It is easy for me to calculate the null space projection matrix P2 = I – B’*inv(BB’)*B. Note that, (xi – yi)’*(xi+yi) = 0 for normalized vectors.

I feel that null(B) = row(A) . Is it true? Or what’s the relationship between null(B) and row(A)? Can I make the conclusion that, if I project data to null(B), those pair of points (xi, yi) will also be maximally separated?

I’m waiting online. Any input will be appreciated! You’re welcome to send me emails!

Thank you very much!

Sue
 
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about null space projection

Homework Statement


If I have a collection of data points (vectors), and x and y are two vectors among them. I want to project the data to a direction that the Euclidean distance between x and y is Maximally preserved. Then this direction should be the row space of (x-y)’, denoted as row( (x-y)’ ), right? (suppose column vectors.)
Now, suppose I have n pairs of such data points grouped as a matrix A, say

A = [
(x1 – y1)’
(x2 – y2)’

(xn –yn)’
]

For my problem, directly calculate the row space projection matrix P1 = A’*inv(AA’)*A is difficult, so I want to do some approximation.



Homework Equations





The Attempt at a Solution



Suppose all the vectors have been normalized, that is norm(x1) = norm(x2) = … = norm(xn) = norm(y1) = .. = norm(yn). I define matrix B as

B = [
(x1 + y1)’
(x2 + y2)’

(xn + yn)’
]

And plan to find the null space of B, denoted as null(B). It is easy for me to calculate the null space projection matrix P2 = I – B’*inv(BB’)*B. Note that, (xi – yi)’*(xi+yi) = 0 for normalized vectors.

I feel that null(B) = row(A) . Is it true? Or what’s the relationship between null(B) and row(A)? Can I make the conclusion that, if I project data to null(B), those pair of points (xi, yi) will also be maximally separated?


I’m waiting online. Any input will be appreciated! You’re welcome to send me emails!

Thank you very much!

Sue
 
Please, give me some feedback
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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