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

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.

Answer 1

[John D'Errico]

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

[Chandan Roy]

Thank you for your answer. I think I need to elaborate my problem in more details.

I need to derive a matrix M which is real and symmetric about its principal diagonal. This M is expressed by

-M = T.L.T' .......(1)

This L has the eigen values of -M. I have derived them from other calculation as follows:

L = diag([-1.702 -0.9962 -0.4424 0.1309 0.7017 1.05 1.654])

Now I need to find T. where T.T'=I.

The first and last rows of T has been determined as follows (this row values are correct):

[-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]

at this point, I need to find the remaining ortogonal rows. As a result, the T would be like this:

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

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.

.

.

.

[0.5784 0.2538 0.2871 0.2870 0.2723 0.1565 0.5822]

After having this T, from equation (1), the final M can be calculated as follows:

[0.0586 -0.0147 -0.2374 -0.0578 0.4314 -0.4385 0]

[-0.0147 -0.0810 0.4825 0.3890 0.6585 0.0952 -1.3957]

[-0.2374 0.4825 0.2431 -0.0022 0.3243 -0.2075 0.1484]

[-0.0578 0.3890 -0.0022 -0.0584 -0.3047 0.4034 -0.0953]

[0.4314 0.6585 0.3243 -0.3047 0.0053 -0.5498 -0.1628]

[-0.4385 0.0952 -0.2075 0.4034 -0.5498 -0.5848 -0.1813]

[0 -1.3957 0.1484 -0.0953 -0.1628 -0.1813 0.0211]

This problem is regarding synthesis of coupling matrix of a 7th order filter. The book didn't explain well. I can't design of own filter without proper understanding on this M derivation and I'm stuck here badly.

Thank you again for your kind help.

[John D'Errico]

It looks like you have now changed things a bit in your elaboration. You now say:

"Now I need to find T. where T.T'=I.

The first and last rows of T has been determined as follows (this row values are correct):

[-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]"

So my first test is to see if you are correct in your statement. Are they orthogonal?

A12 = [-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];
A12*A12'
ans =
       2.2337    -2.58e-06
    -2.58e-06       1.0014

So they are approximately orthogonal. That are not orthonormal.

You have said repeatedly that T*T' will be an identity matrix. But again, I need to state that for your claim to be true, that the diagonal elements of that product needed to be 1. Do I need to syay this a million times? READ WHAT I TELL YOU! The non-zeros in the off-diagonal terms there can be attributed to that you only gave 4 digits in those vectors. But the 2.2337 term is not 1, and not even close.

Sorry. I get frstrated when someone asks a question, but then seems not to be thinking about the answer they get.

While I could orthogonalize A12, the result would NOT be the same. For example, I can use orth to make them an orthonormal set.

A12span = orth(A12.').'
A12span =
    -0.060019      0.44555      -0.5039      0.50484     -0.47781      0.24556     0.022481
     -0.57801     -0.25363      -0.2869     -0.28681     -0.27211      -0.1564     -0.58181

Or, I could just normalize those vectors so they have unit norm. But then they will be different vectors, not as you seem to be asking for.

The rest of your question, once you get a pair of vectors that actually have the properties that you state, is easy to do, but, as I said before, M will not be unique.

Answer 2

[LAKKIMSETTI SUNANDA]

I too have the same question where i have the first and last rows of the matrix(7x7) and i want to find the other 5 rows of the matrix by using Gram Schmidt orthogonalization. Can you help me understand this?

A=[ -0.0600 0.0225 0.2456 0.4455 -0.4777 -0.5041 0.5047 ;

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0.5780 0.5817 0.1563 0.2537 0.2721 0.2869 0.2869 ];

[Chandan Roy]

I solved this 5*5 matrix problem. you can assume rest three rows as I have done in u3, u4 and u5. You can modify these lines code to solve your 7*7 matrix

v1 =A1; %first row elements of a 5*5 A matrix

v2 = A5; %last row elements of a 5*5 A matrix

u3 = [0; 1; 0; 0; 0];

u4 = [0; 0; 1; 0; 0];

u5 = [0; 0; 0; 1; 0];

T3 = u3- (dot(u3,v1)/norm(v1))*v1-(dot(u3,v2)/norm(v2))*v2;

v3 = T3./norm(T3);

T4 = u4- (dot(u4,v1)/norm(v1))*v1-(dot(u4,v2)/norm(v2))*v2-(dot(u4,v3)/norm(v3))*v3;

v4 = T4./norm(T4);

T5 = u5- (dot(u5,v1)/norm(v1))*v1-(dot(u5,v2)/norm(v2))*v2-(dot(u5,v3)/norm(v3))*v3-(dot(u5,v4)/norm(v4))*v4;

v5 = T5./norm(T5);

T = vpa((([v1 v3 v4 v5 v2])'),4); %now you have the complete T matrix

[LAKKIMSETTI SUNANDA]

Thank you but i didn't get the required matrix

i.e., M=

[0.0586 -0.0147 -0.2374 -0.0578 0.4314 -0.4385 0]

[-0.0147 -0.0810 0.4825 0.3890 0.6585 0.0952 -1.3957]

[-0.2374 0.4825 0.2431 -0.0022 0.3243 -0.2075 0.1484]

[-0.0578 0.3890 -0.0022 -0.0584 -0.3047 0.4034 -0.0953]

[0.4314 0.6585 0.3243 -0.3047 0.0053 -0.5498 -0.1628]

[-0.4385 0.0952 -0.2075 0.4034 -0.5498 -0.5848 -0.1813]

[0 -1.3957 0.1484 -0.0953 -0.1628 -0.1813 0.0211]

I have v1=[ -0.0600; 0.0225; 0.2456; 0.4455; -0.4777; -0.5041; 0.5047] ;

v2=[0.5780; 0.5817; 0.1563; 0.2537; 0.2721; 0.2869; 0.2869 ];

These v1 and v2 are normalised by n1 and n2(turns ratio) respectively

and lambda=diag([ -1.7022 1.6543 -0.9963 1.0500 0.7017 -0.4423 0.1309]);

When i find the T matrix with the given formula i'm getting TT'=I but the M matrix is different.

T=[ -0.0599566, 0.0224517, 0.245609, 0.445536, -0.477722, -0.504079, 0.504739]

[ -0.411887, 0.813089, -0.118619, -0.193797, -0.181505, -0.191351, -0.219199]

[ -0.13114, 0, 0.949299, -0.181262, 0.0561061, 0.0592613, -0.205224]

[ -0.273743, 0, 0, 0.816534, 0.145491, 0.153645, -0.462129]

[ -0.266808, 0, 0, 3.4682e-17, 0.800289, -0.473957, 0.252421]

[ -0.573855, 0, 0, 7.2147e-17, 0, 0.612546, 0.543579]

[ 0.578035, 0.581706, 0.156322, 0.253687, 0.27214, 0.28692, 0.286908]

I don't understand where i did the mistake. Can you please help me with this?

[Chandan Roy]

Coupling matrix elements can be different. But you can do some similarity transformations to your coupling matrix M. And then you will find reduced coupling matrix (folded form). However, you can test your coupling matrix M by ploting S-parameters from it. You will get same Scattering matrix from your coupling M even though they are different from your reference. I hade same issue. But, if you get different S-parameters from your reference then you have done mistake in calculation.

[LAKKIMSETTI SUNANDA]

Thank you for your answer.

[Chandan Roy]

My pleasure