MATLAB: If first and last rows of an orthogonal matrix is known, is it possible to determine remaining orthogonal rows

gram-schmidt orthogonalizationorthogonalization

I need to derive a 7*7orthogonal matrix. I have first and last rows of the required matrix. Here is my first and last row:

[-0.0897 0.6659 -0.7531 0.7545 -0.7141 0.3670 0.0336]

[0.5784 0.2538 0.2871 0.2870 0.2723 0.1565 0.5822]

I tried to derive rest of the rows by converting these rows into reduced echolon form as follows:

[1 0 0.9372 0 0.8888 0.0271 0.9294]

[0 1 -1 1.133 -0.9527 0.5548 0.1756]

I end up in following rows

t2 = [-0.9372 1 1 0 0 0 0];

t3 = [0 -1.133 0 1 0 0 0];

t4 = [-0.8888 0.9527 0 0 1 0 0];

t5 = [-0.0271 -0.5548 0 0 0 1 0];

t6 = [-0.9294 -0.1756 0 0 0 0 1];

I suspect, my answer is not correct. In the book, it's written that it can be done by Gram-Schmidt orthogonalization (GSO) process though I couldn't find how to do that. Any hint in order to solve this problem will be appreciated.

Best Answer

If this is homework, and you need to do this using tools other than null, sorry, you are out of luck. I won't do your homework.

Given only TWO rows of a 7x7 orthogonal matrix, the other 5 rows cannot be computed unambiguously. It is trivial to compute 5 rows that are orthogonal to those two. But the solution is not unique.

A12 = [1 0 0.9372 0 0.8888 0.0271 0.9294;0 1 -1 1.133 -0.9527 0.5548 0.1756];

First, these rows are NOT rows of an orthogonal matrix. Not even close. If they were rows of an orthognal matrix, then they would form a 2x2 identity matrix when I did this. (At least the matrix would be diagonal.)

A12*A12'
ans =
       3.5328      -1.6057
      -1.6057         4.53

As you see, not close. We can find a pair of vectors that span the subspace spanned by the rows of A12.

A12span = orth(A12.').'
A12span =
      0.24813     -0.33686      0.56941     -0.38167      0.54147     -0.18017      0.17146
      0.52522      0.38687      0.10536      0.43833     0.098237      0.22887      0.55607
      
A12span*A12span'
ans =
            1   6.1442e-17
   6.1442e-17            1

They form a basi for the given subspace. And as I showed, they are now clearly orthogonal to each other. But, how can you produce 5 rows that are orthogonal to them? That part is trivial, if you understand these tools and linear algebra.

Aetc = null(A12)';

Aetc contains 5 rows that are orthogonal to the first two. It is NOT unique. But Aetc does contain a set of 5 orthogonal vectors, that form the nulllspace of A12.

A12*Aetc'
ans =
  -5.6499e-17  -5.0291e-18   8.9914e-17  -8.8734e-18   1.0304e-16
  -3.4911e-17   1.4416e-16  -4.3053e-17  -3.6445e-17   3.9777e-17

As you can see, these new rows are orthogonal to those we started with (to within floating point trash.) However, as I have said, they are not unique. Here is the set I found from null.

Aetc
Aetc =
     -0.44155      0.37035      0.76564      0.14227     -0.22281     0.066519     -0.08584
    -0.089566     -0.58121      0.11312      0.77673      0.10813     -0.11176     -0.11785
      -0.4184      0.35345     -0.22269      0.13578      0.78829     0.063512    -0.080955
    -0.058912     -0.28873     0.051501     -0.11091     0.049266      0.94435    -0.063196
     -0.53015     -0.23152     -0.11587    -0.088939     -0.10955    -0.049171      0.79346

But I can randomly rotate them. As you can see, they are different. But they still form an orthogonal basis for the nullspace.

Aetc2 = orth(rand(5))*Aetc
Aetc2 =
      0.59296     -0.41116    -0.027373      0.20009    -0.064484      0.34931     -0.55892
      0.25207     -0.19553     -0.49276     -0.61913     -0.19405       0.2677      0.40345
     -0.31595     -0.18661     -0.40325      0.22085      0.71615      0.37217     0.050869
      0.37243      0.44545     -0.42055       -0.177      0.36582      -0.4723     -0.31272
     0.094385     -0.54435     -0.28572      0.39843    -0.093004     -0.59992      0.29299
>> Aetc2*Aetc2'
ans =
            1   1.8385e-16  -2.4559e-16    6.162e-17   -1.085e-16
   1.8385e-16            1   1.9243e-16  -1.2065e-16   2.4558e-16
  -2.4559e-16   1.9243e-16            1   7.8167e-18  -2.8819e-16
    6.162e-17  -1.2065e-16   7.8167e-18            1   1.6765e-16
   -1.085e-16   2.4558e-16  -2.8819e-16   1.6765e-16            1
>> A12*Aetc2'
ans =
   1.6735e-16   1.0458e-17   2.3824e-17   2.5912e-17   3.4096e-17
   3.8011e-17  -6.4272e-18  -3.6084e-17  -9.3707e-17   8.6515e-17

And as shown, they are still orthogonal to the original set in A12.

Finally, IF the goal is to produce an orthogonal 7x7 matrix, this would do so:

A = [orth(A12')';Aetc];

A is an orthogonal matrix, based on the original rows. As a test of that claim...

A*A'
ans =
            1    7.532e-17  -5.9336e-19  -6.0682e-17   4.1989e-17   1.6129e-17    4.086e-17
    7.532e-17            1  -8.3428e-17   5.4361e-17  -2.9411e-17  -2.2233e-17   4.4083e-17
  -5.9336e-19  -8.3428e-17            1   4.1705e-17   4.5634e-17   6.1926e-18   2.3981e-17
  -6.0682e-17   5.4361e-17   4.1705e-17            1   4.5802e-17  -1.7362e-17   7.2233e-19
   4.1989e-17  -2.9411e-17   4.5634e-17   4.5802e-17            1   1.6961e-17  -5.3706e-19
   1.6129e-17  -2.2233e-17   6.1926e-18  -1.7362e-17   1.6961e-17            1   8.1267e-19
    4.086e-17   4.4083e-17   2.3981e-17   7.2233e-19  -5.3706e-19   8.1267e-19            1

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MATLAB: Fourier analysis of waveforms and clipping wave, HELP PLEASE!!!

I'm not sure how you are using the term "period" here. If you are talking about only the first 1 of the fundamental frequency which is 1 MHz here, then I think you want at least a few periods to do a spectral analysis even with a clean waveform. The resolution in the DFT (implemented by fft) depends on the number of samples. The more samples, the better the resolution.

In your case, it's easy to see that

xnew = x(1:2800);

gives you the initial waveform, so why not use this?

     Fs = 1/1e-8;
     d = fdesign.lowpass('Fp,Fst,Ap,Ast',1.2e6,1.4e6,0.5,60,Fs);
     Hd = design(d,'butter');
     xnew = x(1:2800);
     y = filter(Hd,detrend(xnew,0));
     ydft = fft(y);
     freq = 0:Fs/length(y):Fs/2;
     ydft = ydft(1:floor(length(y)/2)+1);
     % Plot in MHz
     plot(freq./1e6,abs(ydft))
     axis([0 10 0 max(abs(ydft))+20])
     xlabel('MHz'); ylabel('Magnitude');

If you copy and paste the above code, it works fine.

MATLAB: Code to produce orthogonal rows in a N* N matrices

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. :)

Best Answer

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