Code to produce orthogonal rows in a N* N matrices

Question

Sudhir Kumar 2020-6-28

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/555940-code-to-produce-orthogonal-rows-in-a-n-n-matrices

编辑： John D'Errico 2020-6-28

采纳的回答： John D'Errico

How to produce N*N matrix where 1st row is ones(1,N) and rest of the rows are orthogonal to each of the row.

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Sudhir Kumar 2020-6-28

No no it can be any number but has to real number thats all

Sudhir Kumar 2020-6-28

Please provide any code if you have in matlab

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Answer 1

John D'Errico 2020-6-28

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https://ww2.mathworks.cn/matlabcentral/answers/555940-code-to-produce-orthogonal-rows-in-a-n-n-matrices#answer_458197

编辑：John D'Errico 2020-6-28

在 MATLAB Online 中打开

Not that hard. First check, is this homework? Hmm. Highly unlikely, since I had to think for at least a second about how to solve it. If your teacher gave you an interesting homework assignment then they caught me. ;-)

I presume the request is to find different, distinct orthogonal matrices every time. Otherwise the answer is trivial. Thus, if you just want the same result every time, then:

% Choose N:
N = 6; 
A = ones(1,N);
A = [A;null(A)'];

The result is N linearly independent orthogonal rows, the first row being all ones. A has the property that A*A' will produce a result indicating the rows are indeed orthogonl. Be careful:

A
A =
            1            1            1            1            1            1
     -0.40825      0.88165     -0.11835     -0.11835     -0.11835     -0.11835
     -0.40825     -0.11835      0.88165     -0.11835     -0.11835     -0.11835
     -0.40825     -0.11835     -0.11835      0.88165     -0.11835     -0.11835
     -0.40825     -0.11835     -0.11835     -0.11835      0.88165     -0.11835
     -0.40825     -0.11835     -0.11835     -0.11835     -0.11835      0.88165
>> A*A'
ans =
            6  -8.3267e-17  -1.1102e-16  -8.3267e-17  -4.1633e-17            0
  -8.3267e-17            1   1.3046e-17   1.3046e-17   7.8421e-18     1.18e-18
  -1.1102e-16   1.3046e-17            1   2.6924e-17   7.8421e-18     1.18e-18
  -8.3267e-17   1.3046e-17   2.6924e-17            1   7.8421e-18     1.18e-18
  -4.1633e-17   7.8421e-18   7.8421e-18   7.8421e-18            1     1.18e-18
            0     1.18e-18     1.18e-18     1.18e-18     1.18e-18            1

As you can see, because you insisted the first row must be all ones, it has not been made to have unit 2-norm. There for the (1,1) element of A*A' wil not be 1, but always N. Worse, A'*A will not be even close to an identity. But you insisted the matrix have a unit first row. So if I then normalize the first row of A...

A(1,:) = A(1,:)/norm(A(1,:))
A =
      0.40825      0.40825      0.40825      0.40825      0.40825      0.40825
     -0.40825      0.88165     -0.11835     -0.11835     -0.11835     -0.11835
     -0.40825     -0.11835      0.88165     -0.11835     -0.11835     -0.11835
     -0.40825     -0.11835     -0.11835      0.88165     -0.11835     -0.11835
     -0.40825     -0.11835     -0.11835     -0.11835      0.88165     -0.11835
     -0.40825     -0.11835     -0.11835     -0.11835     -0.11835      0.88165
>> A'*A
ans =
            1   5.6919e-17   2.9163e-17   5.6919e-17   9.8552e-17   1.2319e-16
   5.6919e-17            1   1.3046e-17   1.3046e-17   7.8421e-18     1.18e-18
   2.9163e-17   1.3046e-17            1   2.6924e-17   7.8421e-18     1.18e-18
   5.6919e-17   1.3046e-17   2.6924e-17            1   7.8421e-18     1.18e-18
   9.8552e-17   7.8421e-18   7.8421e-18   7.8421e-18            1     1.18e-18
   1.2319e-16     1.18e-18     1.18e-18     1.18e-18     1.18e-18            1
>> A*A'
ans =
            1  -5.6919e-17  -2.9163e-17  -5.6919e-17  -9.8552e-17  -1.2319e-16
  -5.6919e-17            1   1.3046e-17   1.3046e-17   7.8421e-18     1.18e-18
  -2.9163e-17   1.3046e-17            1   2.6924e-17   7.8421e-18     1.18e-18
  -5.6919e-17   1.3046e-17   2.6924e-17            1   7.8421e-18     1.18e-18
  -9.8552e-17   7.8421e-18   7.8421e-18   7.8421e-18            1     1.18e-18
  -1.2319e-16     1.18e-18     1.18e-18     1.18e-18     1.18e-18            1

Now both products are essentially identity matrices to within floating point trash. The problem is, this result is not random in any sense. And I gather you wanted random. That is also not difficult, and a simple extension. Do you see how? The basic trick is still the same. Still not hard.

N = 4;
A = ones(1,N);
V2 = randn(1,N); % a random start
V2 = V2/norm(V2);
V2 = V2 - A*dot(V2,A)/norm(V2)/N;
A = [A;V2/norm(V2)];
A = [A;null(A)'];

This yields a different matrix each time with the desired property. Note that again, you needed to scale the first row to have unit norm if you will try to form A'*A to test. Here, with N == 4, I get the expected result, and it will be a new such matrix every time, because the SECOND row is chosen randomly.

A
A =
            1            1            1            1
      0.63227     -0.59478       0.3318     -0.36929
      -0.5759     -0.07093      0.79992     -0.15308
     -0.13625     -0.62546   -0.0065307      0.76824
     
A*A'
ans =
            4   2.2204e-16   2.7756e-16   5.5511e-17
   2.2204e-16            1   4.1633e-17   4.1633e-17
   2.7756e-16   4.1633e-17            1  -1.3878e-17
   5.5511e-17   4.1633e-17  -1.3878e-17            1

As I said, easy enough to solve. The trick is to choose one new vector as the second row randomly. Then employ one step of a Gram-Schmidt process to make it orthogonal to the first row. Then use null to produce the other N-2 rows.

https://en.wikipedia.org/wiki/Gram–Schmidt_process

Just thinking, that points out an obvious alternative, but it is no better really. That is, I could have generated N-1 random rows. Then subtract off the component of each row that has non-zero projection on the vector ones(1,N). Finally, use orth to make those N-1 rows orthongonal to each other. As I said. No better, and probably slightly slower to run for large N.

If it was homeowrk, then I apologize to your teacher. :)