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From: Bruno Luong on 25 Apr 2010 17:59

> -----------------------------
> By using of orthogonal polynomials , I take only coefficients for matrix T.
> But , when I want to construct above matrix Q, I don't obtained
> V*Q'=zeros(4)
> Q*V'=zeros(4)
> V*V'=eye(4)
> that is .
> best regards.
>

From HELP NULL (Thanks to Roger who has corrected me by pointing to NULL, which I always mistaken with ORTH for some reason):

NULL Null space.
Z = NULL(A) is an orthonormal basis for the null space of A obtained
from the singular value decomposition. That is, A*Z has negligible
elements, size(Z,2) is the nullity of A, and Z'*Z = I.

Don't you see the relation?

Bruno
From: Roger Stafford on 25 Apr 2010 18:58
"atanas " <atanaslove2000(a)abv.bg> wrote in message <hr2acs$mgu$1(a)fred.mathworks.com>...
> By using of orthogonal polynomials , I take only coefficients for matrix T.
> ........
----------------
Constructing a T matrix involves, in my opinion, finding an orthonormal basis in R^4 in which the first two basis elements are destined to lie the 2D space spanned by the two rows of [C0 C1] and the second two basis elements lie in the 2D space spanned by the two rows of [C2 C3]. That way you will be assured that the two rows of [C0 C1] will be orthogonal to the two rows of [C2 C3]. By applying this process to the two rows of [C0 C1 C2 C3] you can eventually determine a suitable T matrix. Finding such an orthonormal set is why Bruno and I keep mentioning the 'null' and 'orth' matlab functions.

How is it you see orthogonal polynomials being useful for creating T? Are you willing to give the details of this?

Roger Stafford
From: atanas on 25 Apr 2010 19:28
"Bruno Luong" <b.luong(a)fogale.findmycountry> wrote in message <hr2dv8$ali$1(a)fred.mathworks.com>...
>
> > -----------------------------
> > By using of orthogonal polynomials , I take only coefficients for matrix T.
> > But , when I want to construct above matrix Q, I don't obtained
> > V*Q'=zeros(4)
> > Q*V'=zeros(4)
> > V*V'=eye(4)
> > that is .
> > best regards.
> >
>
> From HELP NULL (Thanks to Roger who has corrected me by pointing to NULL, which I always mistaken with ORTH for some reason):
>
> NULL Null space.
> Z = NULL(A) is an orthonormal basis for the null space of A obtained
> from the singular value decomposition. That is, A*Z has negligible
> elements, size(Z,2) is the nullity of A, and Z'*Z = I.
>
> Don't you see the relation?
>
> Bruno

Sir Bruno,
thank you reply,
Yes, I see the relation,BUT!
- I don't have matrix A. Which is a matrix must be estimation with Z=null(?) ?
- If you speech for matrix T, that is easy to find its orthonal matrix! I knew this.
- BUT, matrix T, go to matrix V (above)
V=[T zeros(2,4),
zeros(2,4) T];
and I have result
T*T'=eye(2) !
and
V*V=eye(4) !
In thi case, I can't used matlab function orth!
If obtained Z=otrh(T), I can't obtanied V*V'=eye(4) and V*Q'=zeros(4)!
If your have solutions, I hope to learning from your.
I am student and I don't know everything.
Thank sir Bruno. That is already.
From: Bruno Luong on 26 Apr 2010 04:25
"atanas " <atanaslove2000(a)abv.bg> wrote in message <hr2j62$djq$1(a)fred.mathworks.com>...

> Yes, I see the relation...!

Obviously you don't, if you see the relation the solution is already found.

NULL(A) returns a orthogonal matrix where each column is orthogonal to all the row of A.

What do the following conditions mean?

> V*Q'=zeros(2); (condition 2)
> Q*V'=zeros(2); (condition 3)
> Q*Q'=eye(2); (condition 4).

Conditions (3) and (2) state the same thing, because (V*Q')' = Q*V'.

Bruno
From: Roger Stafford on 26 Apr 2010 13:35
"atanas " <atanaslove2000(a)abv.bg> wrote in message <hqvsda$m6d$1(a)fred.mathworks.com>...
> Hello,
> I have problem for my project:
> Let we matrices C0,C1,C2,C3,D0,D1,D2, and D3 are with size 2x2.
> We knowing matrices C0,C1,C2,and C3 that
> [C0 C1 C2 C3]*[C0 C1 C2 C3]'=I
> We construct the matrix
> W=[C0 C1 C2 C3 zeros(2,2) zeros(2,2);
> zeros(2,2) zeros(2,2) C0 C1 C2 C3;
> D0 D1 D2 D3 zeros(2,2) zeros(2,2);
> zeros(2,2) zeros(2,2) D0 D1 D2 D3].
>
> How to find matrix D0,D1,D2, and D3 so that satisfy condition W*W'=I?
-----------------
Atanas, this thread is stretching out too long, so in spite of what I said earlier, I have decided to give you my version of a matlab code that would solve your problem. I hope you will find it useful. I leave it up to you to discover the logic behind the algorithm.

Remember, it is very important that any set of C0, C1, C2, C3 you create yourself instead of using part 1 should pass the test that is given there with just as much accuracy as in part 1, namely with errors only out in the 14th or 15th decimal place. Otherwise the procedure in part 2 will not be accurate. It depends on the C's being correctly chosen.

If you recall, I mentioned earlier that there is one degree of freedom in the choice of the D's, for any given set of C's. In the code below in the line

y = null([y1,y2])';

matlab's 'null' function must choose two normal mutually orthogonal four-element vectors which are orthogonal to y1 and y2, so it must choose them in a two-dimensional subspace. However, the two vectors could be at any rotated orientation in this subspace and this is where the one degree of freedom for the D's comes in.

Roger Stafford
--------------------------
% Part 1 - Random generation of C's
clear
x = orth(randn(4))';
y = orth(randn(4,2))';
C0 = y(:,1:2)*x(1:2,1:2);
C1 = y(:,1:2)*x(1:2,3:4);
C2 = y(:,3:4)*x(3:4,1:2);
C3 = y(:,3:4)*x(3:4,3:4);

% Check the C's
Z2 = zeros(2);
V = [C0,C1,C2,C3,Z2,Z2;
Z2,Z2,C0,C1,C2,C3];
format long
disp(max(max(abs(V*V'-eye(4)))))
clear V Z2 x y

% Part 2 - Given the proper C's, create D's
x1 = orth([C0,C1]')';
x2 = orth([C2,C3]')';
y1 = [C0,C1]/x1;
y2 = [C2,C3]/x2;
y = null([y1,y2])';
D0 = y(:,1:2)*x1(:,1:2);
D1 = y(:,1:2)*x1(:,3:4);
D2 = y(:,3:4)*x2(:,1:2);
D3 = y(:,3:4)*x2(:,3:4);

% Check the D's
Z2 = zeros(2);
W = [C0,C1,C2,C3,Z2,Z2;
Z2,Z2,C0,C1,C2,C3;
D0,D1,D2,D3,Z2,Z2;
Z2,Z2,D0,D1,D2,D3];
format long
disp(max(max(abs(W*W'-eye(8)))))
clear W Z2 x1 x2 y y1 y2
--------------------------
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