How to generate random projection matrices?

As it is said in the question, I am looking for a Matlab function that generates random projection matrices, so that I can use it for linear programming.

Answer 1

[KALYAN ACHARJYA]

function P=projection_mat(n)
A=colbasis(magic(n));
P=A*inv(A'*A)*A';
end

The colbasis function is here

Here n represent size of square matrix. Please note that I have answered this question from here

Command Window:

>> y=projection_mat(6)
y =
    0.7500   -0.0000    0.2500    0.2500   -0.0000   -0.2500
   -0.0000    1.0000    0.0000   -0.0000   -0.0000    0.0000
    0.2500    0.0000    0.7500   -0.2500   -0.0000    0.2500
    0.2500   -0.0000   -0.2500    0.7500   -0.0000    0.2500
   -0.0000   -0.0000   -0.0000   -0.0000    1.0000   -0.0000
   -0.2500    0.0000    0.2500    0.2500   -0.0000    0.7500

You can generate any size matries, just pass the same size matrix to colbasis function.

Hope it helps!

[Rufat Imanov]

Thanks sir

[Rufat Imanov]

I have a question, the function generates only the same matrices, so it does not generate random matrix, but generates a single solution

[KALYAN ACHARJYA]

Is there any necessity having fixed size matrices?

>> y=projection_mat(6)
y =
    0.7500   -0.0000    0.2500    0.2500   -0.0000   -0.2500
   -0.0000    1.0000    0.0000   -0.0000   -0.0000    0.0000
    0.2500    0.0000    0.7500   -0.2500   -0.0000    0.2500
    0.2500   -0.0000   -0.2500    0.7500   -0.0000    0.2500
   -0.0000   -0.0000   -0.0000   -0.0000    1.0000   -0.0000
   -0.2500    0.0000    0.2500    0.2500   -0.0000    0.7500
>> y=projection_mat(5)
y =

    1.0000   -0.0000   -0.0000   -0.0000   -0.0000
   -0.0000    1.0000   -0.0000   -0.0000   -0.0000
   -0.0000   -0.0000    1.0000   -0.0000    0.0000
   -0.0000   -0.0000   -0.0000    1.0000    0.0000
   -0.0000   -0.0000   -0.0000   -0.0000    1.0000

>> 

[Rufat Imanov]

yes, the size of matrix should be the same, but the matrix by itslef needs to change. Beacuse, I am working on finding a specific matrix that can project a square into square, so that the nodes/edges are not outside of the boundary of the square.

Answer 2

[Bruno Luong]

n = 5
r = 3; % rank, dimension of the projection subspace

[Q,~] = qr(randn(n));
Q = Q(:,1:r);
P = Q*Q' % random projection matrix P^2 = P, rank P = r

[Rufat Imanov]

Are you sure that your function produces a projection matrix, because when I ran it in my pc, I found out that the dot ptoduct of any pair of rows or columns is not 0. Also, R^3 = R

[Bruno Luong]

Yes I'm sure of P^2 = P and rank(P) is as specified. That's pretty much a projection by definition.

Actually P^n = P for any n >= 1 (you project an object n time, ot does not change after the first time).

I don't know why your expect columns (or rows, since P is symmetric) are orthogonal. The projection of basic vectors is in general NOT orthogonal (e.g., the 3D cube drawed on your computer screen does not have right angles in general).

Checkout https://en.wikipedia.org/wiki/Projection_matrix for property.

n = 5
r = 3; % rank, dimension of the projection subspace
[Q,~] = qr(randn(n));
Q = Q(:,1:r);
P = Q*Q' % random projection matrix P^2 = P, rank P = r

Outputs:

n =

     5


P =

    0.3071   -0.1744   -0.3665   -0.0818   -0.2034
   -0.1744    0.9371   -0.1236   -0.1015    0.0542
   -0.3665   -0.1236    0.7542   -0.1770    0.0667
   -0.0818   -0.1015   -0.1770    0.6461    0.4247
   -0.2034    0.0542    0.0667    0.4247    0.3555

>> P^2

ans =

    0.3071   -0.1744   -0.3665   -0.0818   -0.2034
   -0.1744    0.9371   -0.1236   -0.1015    0.0542
   -0.3665   -0.1236    0.7542   -0.1770    0.0667
   -0.0818   -0.1015   -0.1770    0.6461    0.4247
   -0.2034    0.0542    0.0667    0.4247    0.3555

>> P

P =

    0.3071   -0.1744   -0.3665   -0.0818   -0.2034
   -0.1744    0.9371   -0.1236   -0.1015    0.0542
   -0.3665   -0.1236    0.7542   -0.1770    0.0667
   -0.0818   -0.1015   -0.1770    0.6461    0.4247
   -0.2034    0.0542    0.0667    0.4247    0.3555

>> norm(P^2-P)

ans =

   5.8312e-16

>> rank(P)

ans =

     3

[Rufat Imanov]

You are completely right, I have miststated my question, I need to generate an orthogonal projection matrix

[Bruno Luong]

Sorry I think the only projection matrix that is orthogonal is diagonal matrix with 1 or 0 on the diagonal. So there is no really randomness for what you ask.

[Bruno Luong]

I wonder if you mistaken "orthogonal projection matrix" and "projection matrix that is orthogonal". They are not the same.

Mine is "orthogonal projection matrix", which is projection matrix (P^2==P) that has additional properties

1. symmetric
2. all eigen values are 0 or 1.

Answer 3

[Image Analyst]

Not sure what you mean by projection, but the radon transform does projections. That's its claim to fame. It basically projects a matrix along any angle and gives you the sum of the interpolated values along the projection angle. This is the crucial function for reconstructing 3-D volumetric CT images from 2-D projections.

The radon() function requires the Image Processing Toolbox.