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From: atanas on 25 Apr 2010 13:43
"Roger Stafford" <ellieandrogerxyzzy(a)mindspring.com.invalid> wrote in message <hr1tgd$fe5$1(a)fred.mathworks.com>...
> "atanas " <atanaslove2000(a)abv.bg> wrote in message <hr1p7k$dtd$1(a)fred.mathworks.com>...
> > Sir Stafford,
> > thank against.
> > C's matrices are ready orhogonal.
> > D's matrices need to find that
> > Q = [D0 D1 D2 D3 zeros(2,2) zeros(2,2);
> > zeros(2,2) zeros(2,2) D0 D1 D2 D3]
> > must satisfy
> > V*V'=eye(2); (condition 1)
> > V*Q'=zeros(2); (condition 2)
> > Q*V'=zeros(2); (condition 3)
> > Q*Q'=eye(2); (condition 4).
> > I have already only condition 1!
> > Need to find conditions 2, 3 and 4.
> > That all.
> > Thanks for your help.
> ----------
> That answers my first question, but my second question was "Have you figured out how to generate examples of C's in matlab that would satisfy (2)?" That is, V*V' = eye(4). Never mind the D's for the moment.
>
> As Bruno indicates, you can make good use of 'orth'. The same is true of 'null'.
>
> Have a try at it, Atanas. I think you will find the effort very rewarding.
>
> Roger Stafford

Sir Stafford,
This is matrix L:
V =[ 0 -0.0407 0.0873 -0.1484 0.9072 0.3460 0.1610 0
0 0.0048 -0.0103 0.0175 -0.1069 -0.0998 0.8401 0.5219];
>> V*V'
ans =
1.0000 0.0000
0.0000 1.0000
This is good!
I need to find matrix Q!
Thanks again.
From: atanas on 25 Apr 2010 14:05
"atanas " <atanaslove2000(a)abv.bg> wrote in message <hr1uvb$iel$1(a)fred.mathworks.com>...
> "Roger Stafford" <ellieandrogerxyzzy(a)mindspring.com.invalid> wrote in message <hr1tgd$fe5$1(a)fred.mathworks.com>...
> > "atanas " <atanaslove2000(a)abv.bg> wrote in message <hr1p7k$dtd$1(a)fred.mathworks.com>...
> > > Sir Stafford,
> > > thank against.
> > > C's matrices are ready orhogonal.
> > > D's matrices need to find that
> > > Q = [D0 D1 D2 D3 zeros(2,2) zeros(2,2);
> > > zeros(2,2) zeros(2,2) D0 D1 D2 D3]
> > > must satisfy
> > > V*V'=eye(2); (condition 1)
> > > V*Q'=zeros(2); (condition 2)
> > > Q*V'=zeros(2); (condition 3)
> > > Q*Q'=eye(2); (condition 4).
> > > I have already only condition 1!
> > > Need to find conditions 2, 3 and 4.
> > > That all.
> > > Thanks for your help.
> > ----------
> > That answers my first question, but my second question was "Have you figured out how to generate examples of C's in matlab that would satisfy (2)?" That is, V*V' = eye(4). Never mind the D's for the moment.
> >
> > As Bruno indicates, you can make good use of 'orth'. The same is true of 'null'.
> >
> > Have a try at it, Atanas. I think you will find the effort very rewarding.
> >
> > Roger Stafford
>
> Sir Stafford,
> This is matrix L:
> V =[ 0 -0.0407 0.0873 -0.1484 0.9072 0.3460 0.1610 0
> 0 0.0048 -0.0103 0.0175 -0.1069 -0.0998 0.8401 0.5219];
V*V' =eye(2)
and
[V zeros(2,4); zeros(2,4) V ][V zeros(2,4); zeros(2,4) V ]'=eye(4) !
This is good!
I need to find matrix Q!
Thanks again.
From: atanas on 25 Apr 2010 14:19
"atanas " <atanaslove2000(a)abv.bg> wrote in message <hr2095$c8s$1(a)fred.mathworks.com>...
> "atanas " <atanaslove2000(a)abv.bg> wrote in message <hr1uvb$iel$1(a)fred.mathworks.com>...
> > "Roger Stafford" <ellieandrogerxyzzy(a)mindspring.com.invalid> wrote in message <hr1tgd$fe5$1(a)fred.mathworks.com>...
> > > "atanas " <atanaslove2000(a)abv.bg> wrote in message <hr1p7k$dtd$1(a)fred.mathworks.com>...
> > > > Sir Stafford,
> > > > thank against.
> > > > C's matrices are ready orhogonal.
> > > > D's matrices need to find that
> > > > Q = [D0 D1 D2 D3 zeros(2,2) zeros(2,2);
> > > > zeros(2,2) zeros(2,2) D0 D1 D2 D3]
> > > > must satisfy
> > > > V*V'=eye(2); (condition 1)
> > > > V*Q'=zeros(2); (condition 2)
> > > > Q*V'=zeros(2); (condition 3)
> > > > Q*Q'=eye(2); (condition 4).
> > > > I have already only condition 1!
> > > > Need to find conditions 2, 3 and 4.
> > > > That all.
> > > > Thanks for your help.
> > > ----------
> > > That answers my first question, but my second question was "Have you figured out how to generate examples of C's in matlab that would satisfy (2)?" That is, V*V' = eye(4). Never mind the D's for the moment.
> > >
> > > As Bruno indicates, you can make good use of 'orth'. The same is true of 'null'.
> > >
> > > Have a try at it, Atanas. I think you will find the effort very rewarding.
> > >
> > > Roger Stafford
> >
> > Sir Stafford,
This is matrix T:
T =[ 0 -0.0407 0.0873 -0.1484 0.9072 0.3460 0.1610 0
0 0.0048 -0.0103 0.0175 -0.1069 -0.0998 0.8401 0.5219];
satisfy T*T' =eye(2) !
Now this matrix T I insert i matrix V
V=[T zeros(2,4);
zeros(2,4) T ];
and
V*V'=eye(4)!
This is correct.
Now, I need to find matrix
Q=[ M zeros(2,4);
zeros(2,4) M ];
must satisfy
V*Q'=zeros(4)
Q*V'=zeros(4)
V*V'=eye(4)
For this reason I must to find submatrix M, such that
M*M'=eye(2) and its V*V'=eye(4)!
My be this is clear.
Thanks.
This is good!
I need to find matrix Q!
From: Roger Stafford on 25 Apr 2010 15:09
"atanas " <atanaslove2000(a)abv.bg> wrote in message <hr212p$38m$1(a)fred.mathworks.com>...
> This is matrix T:
> T =[ 0 -0.0407 0.0873 -0.1484 0.9072 0.3460 0.1610 0
> 0 0.0048 -0.0103 0.0175 -0.1069 -0.0998 0.8401 0.5219];
> satisfy T*T' =eye(2) !
> Now this matrix T I insert i matrix V
> V=[T zeros(2,4);
> zeros(2,4) T ];
> and
> V*V'=eye(4)!
> This is correct.
> Now, I need to find matrix
> Q=[ M zeros(2,4);
> zeros(2,4) M ];
> must satisfy
> V*Q'=zeros(4)
> Q*V'=zeros(4)
> V*V'=eye(4)
> For this reason I must to find submatrix M, such that
> M*M'=eye(2) and its V*V'=eye(4)!
> My be this is clear.
> Thanks.
> This is good!
> I need to find matrix Q!
--------------
You have shown me a matrix T (a.k.a. matrix L) that works (approximately), but you haven't shown me how you got it. It is in the process of obtaining T wherein useful things happen in your problem. Just for now forget all about D and Q, and concentrate on the steps involved in finding a successful T. You didn't get it by simply finding two random eight-element vectors for the rows of [C0 C1 C2 C3] which are orthogonal and normalizing them. There is more to it than that, because you will also have to make the two four-element rows of [C0 C1] both orthogonal to the two four-element rows of [C2 C3] which would not necessarily be the case (though it appears to be true of the example you have shown.) Again the question is, how did you get your successful T?

Roger Stafford
From: atanas on 25 Apr 2010 16:58
"Roger Stafford" <ellieandrogerxyzzy(a)mindspring.com.invalid> wrote in message <hr240l$8b8$1(a)fred.mathworks.com>...
> "atanas " <atanaslove2000(a)abv.bg> wrote in message <hr212p$38m$1(a)fred.mathworks.com>...
> > This is matrix T:
> > T =[ 0 -0.0407 0.0873 -0.1484 0.9072 0.3460 0.1610 0
> > 0 0.0048 -0.0103 0.0175 -0.1069 -0.0998 0.8401 0.5219];
> > satisfy T*T' =eye(2) !
> > Now this matrix T I insert i matrix V
> > V=[T zeros(2,4);
> > zeros(2,4) T ];
> > and
> > V*V'=eye(4)!
> > This is correct.
> > Now, I need to find matrix
> > Q=[ M zeros(2,4);
> > zeros(2,4) M ];
> > must satisfy
> > V*Q'=zeros(4)
> > Q*V'=zeros(4)
> > V*V'=eye(4)
> > For this reason I must to find submatrix M, such that
> > M*M'=eye(2) and its V*V'=eye(4)!
> > My be this is clear.
> > Thanks.
> > This is good!
> > I need to find matrix Q!
> --------------
> You have shown me a matrix T (a.k.a. matrix L) that works (approximately), but you haven't shown me how you got it. It is in the process of obtaining T wherein useful things happen in your problem. Just for now forget all about D and Q, and concentrate on the steps involved in finding a successful T. You didn't get it by simply finding two random eight-element vectors for the rows of [C0 C1 C2 C3] which are orthogonal and normalizing them. There is more to it than that, because you will also have to make the two four-element rows of [C0 C1] both orthogonal to the two four-element rows of [C2 C3] which would not necessarily be the case (though it appears to be true of the example you have shown.) Again the question is, how did you get your successful T?
>
> Roger Stafford
-----------------------------
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.

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