How to create an orthogonal matrix in Matlab with one fixed column

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I want to create a square NxN matrix orthogonal, with the constraint that the first column is a column vector of k*ones(N,1), where k is a constant at choice. Is there any procedure?

I.e.

A= [k * *;k * *;k * *]

is a 3x3 matrix, where the first column is a vector k*ones(3,1), and the other two vectors have to be created in such a way the matrix is orthogonal

Maybe this should be posted on Math.StackEchange where equations can be typed properly if you want a proper theoretical explanation. But if just want the code...

First, if N is the dimension of your matrix, this constrains the value of k to:

k=sqrt(1/N);
A(1,:) = k*ones(1,N);

Then, the second row can be constructed with:

A(2,:) = sqrt(0.5)*[1,-1,zeros(1,N-2)];

This creates a simple vector orthogonal to the first.

The third line can be computed with:

aux = [1,1,-2,zeros(1,N-3)];
A(3,:) = aux/norm(aux);

The fourth:

aux = [1,1,1,-3,zeros(1,N-4)];
A(4,:) = aux/norm(aux);

And so on.

In short:

A=zeros(N);
k=sqrt(1/N);
A(1,:) = k*ones(1,N);
for i=2:N
    aux = [ones(1,i-1),-(i-1),zeros(1,N-i)];
    A(i,:) = aux/norm(aux);
end

orthogonal matrix with different size of rows and columns, I want to create a square NxN matrix orthogonal, with the constraint that the first column is a column vector of k*ones(N,1), where k is a constant at choice. Expanding a Matrix. You can add one or more elements to a matrix by placing them outside of the existing row and column index boundaries. MATLAB automatically pads the matrix with zeros to keep it rectangular. For example, create a 2-by-3 matrix and add an additional row and column to it by inserting an element in the (3,4) position.

Alternatively:

kk = k*ones(3,1); % fixed first column (could be anything)
X = [kk, null(kk')']

gives you a square matrix with mutually orthogonal columns, no matter what's the vector kk. It will be an orthonormal matrix only when norm(k)==1 (which implies k=1/sqrt(3) in your examples, as the others have noted). Note that the first case does not imply its rows are orthogonal, whereas the second one does.

How to create an orthogonal basis for the column space of a matrix , how to generate orthogonal matrix with different size of rows and columns? If your goal is to find a set of n vectors of length d, such that they are orthogonal to each The answer is as simple as virtually one line of code. In logical indexing, you use a single, logical array for the matrix subscript. MATLAB extracts the matrix elements corresponding to the nonzero values of the logical array. The output is always in the form of a column vector. For example, A(A > 12) extracts all the elements of A that are greater than 12.

I'm sorry I can't quote MATLAB code, but I'm sure this will be straightforward to code.

What you want is a Householder reflector These are othonormal matrices that are symmetric (and so are their own inverses). Given a vector v you can find a reflector H so that

H*v = a*e1

where a is += length of v and e1 is (1,0,..)'

I'd imagine MATLAB has a routine for this.

Given the properties of H this means that the first column of H is parallel to v, and the other columns are orthogonal to v and to each other.

So if you scale the first column of H (it is of length 1) appropriately, you have your desired matrix.

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How do I create orthogonal basis based on two "almost", As in case of order 3 I need two orthogonal matrices with fixed hamming distance 3 No number from {1,2,3} is repeated in row , column repetition is allow. I have a 469x1 matrix (interNode) and I want to make a second column with all rows having the same value (2) to give me a 469x2 matrix. Any help would be appreciated :) 0 Comments

Comments
  • Could you please provide a small example of what you mean (input and output)?
  • I have edited the question: input the first column vector, output the resulting orthogonal matrix
  • This is a right one solution. Thanks! This procedure has a name or is an empirical one?
  • I'm not aware of a name or a reference to it.
  • Thank you. I think this is a valid solution for the problem! :)