boguS <bteodoru@gmail.com> wrote in message
<f9d38db3-ae05-4620-a991-46f61d70453c@t54g2000hsg.googlegroups.com>...
> On 27 Iun, 18:18, boguS <bteod...@gmail.com> wrote:
> > For two known matrices [k] and [k1] I need to solve
<lambda> from Eq:
> >
> > [k1]=lambda*[k]
>
> <lambda> is scalar.
Uhh, can't you just divide any element of k1 by the
corresponding element in k? Assuming the elements aren't
zero, that is...
On 27 Iun, 19:29, "Ian Clarkson" <ovoidkumq...@hotmail.com> wrote:
> boguS <bteod...@gmail.com> wrote in message
>
> <f9d38db3-ae05-4620-a991-46f61d704...@t54g2000hsg.googlegroups.com>...
>
> > On 27 Iun, 18:18, boguS <bteod...@gmail.com> wrote:
> > > For two known matrices [k] and [k1] I need to solve
> <lambda> from Eq:
>
> > > [k1]=lambda*[k]
>
> > <lambda> is scalar.
>
> Uhh, can't you just divide any element of k1 by the
> corresponding element in k? Assuming the elements aren't
> zero, that is...
No, it's not so simple...because I need to write the known [k1] matrix
in such way that it must be <lambda*[k]> were [k] is just a (known)
matrix.
>> > On 27 Iun, 18:18, boguS <bteod...@gmail.com> wrote:
>> > > For two known matrices [k] and [k1] I need to solve
>> <lambda> from Eq:
>> > > [k1]=lambda*[k]
>> > <lambda> is scalar.
>> Uhh, can't you just divide any element of k1 by the
>> corresponding element in k? Assuming the elements aren't
>> zero, that is...
>No, it's not so simple...because I need to write the known [k1] matrix
>in such way that it must be <lambda*[k]> were [k] is just a (known)
>matrix.
Could you give an example? With the way you have stated your problem,
Ian's response appears correct to me.
Is the following code not an implementation of the definition
of lambda (a scalar) times k (a matrix) ?
k2 = zeros(size(k));
for idx1 = 1:size(k,1)
for idx2 = 1:size(k,2)
k2(i,j) = lambda * k(idx1,idx2);
end
end
If I haven't forgotten my linear algebra too badly, then that
would mean that k1 would have to be the same size as k, and
that L = k1 ./ k would have to be a matrix with all entries equal
to lambda (except for those entries where k was 0.) So dividing
any one k1 entry by its corresponding non-zero k entry would
recover lambda.
--
"This quitting thing, it's a hard habit to break once you start."
-- Walter Matthau
boguS <bteodoru@gmail.com> wrote in message
<7973ad19-75f0-4c31-8f1e-
c786818e1e63@z72g2000hsb.googlegroups.com>...
> No, it's not so simple...because I need to write the known [k1] matrix
> in such way that it must be <lambda*[k]> were [k] is just a (known)
> matrix.
Well, you can perform some fancy manipulations like this:
a = trace(k'*k);
b = trace(k'*k1);
c = trace(k1'*k1);
lambda = b/a;
Then if lambda^2*a + 2*lambda*b + c is equal to zero, you have your solution.
Otherwise, there can be no exact solution, but this lambda does give the least
squares approximation of lambda*k-k1 to zero.
"Roger Stafford" <ellieandrogerxyzzy@mindspring.com.invalid> wrote in
message <g438v1$keu$1@fred.mathworks.com>...
> boguS <bteodoru@gmail.com> wrote in message
> <7973ad19-75f0-4c31-8f1e-
> c786818e1e63@z72g2000hsb.googlegroups.com>...
> > No, it's not so simple...because I need to write the known [k1] matrix
> > in such way that it must be <lambda*[k]> were [k] is just a (known)
> > matrix.
>
> Well, you can perform some fancy manipulations like this:
>
> a = trace(k'*k);
> b = trace(k'*k1);
> c = trace(k1'*k1);
> lambda = b/a;
>
> Then if lambda^2*a + 2*lambda*b + c is equal to zero, you have your
solution.
> Otherwise, there can be no exact solution, but this lambda does give the
least
> squares approximation of lambda*k-k1 to zero.
>
> Roger Stafford
>
Why so much effort? The least squares solution
is just this:
On 27 Iun, 20:27, rober...@ibd.nrc-cnrc.gc.ca (Walter Roberson) wrote:
> In article <7973ad19-75f0-4c31-8f1e-c786818e1...@z72g2000hsb.googlegroups.com>,
>
> boguS <bteod...@gmail.com> wrote:
> >On 27 Iun, 19:29, "Ian Clarkson" <ovoidkumq...@hotmail.com> wrote:
> >> boguS <bteod...@gmail.com> wrote in message
> >> <f9d38db3-ae05-4620-a991-46f61d704...@t54g2000hsg.googlegroups.com>...
> >> > On 27 Iun, 18:18, boguS <bteod...@gmail.com> wrote:
> >> > > For two known matrices [k] and [k1] I need to solve
> >> <lambda> from Eq:
> >> > > [k1]=lambda*[k]
> >> > <lambda> is scalar.
> >> Uhh, can't you just divide any element of k1 by the
> >> corresponding element in k? Assuming the elements aren't
> >> zero, that is...
> >No, it's not so simple...because I need to write the known [k1] matrix
> >in such way that it must be <lambda*[k]> were [k] is just a (known)
> >matrix.
>
> Could you give an example? With the way you have stated your problem,
> Ian's response appears correct to me.
>
> Is the following code not an implementation of the definition
> of lambda (a scalar) times k (a matrix) ?
>
> k2 = zeros(size(k));
> for idx1 = 1:size(k,1)
> for idx2 = 1:size(k,2)
> k2(i,j) = lambda * k(idx1,idx2);
> end
> end
>
> If I haven't forgotten my linear algebra too badly, then that
> would mean that k1 would have to be the same size as k, and
> that L = k1 ./ k would have to be a matrix with all entries equal
> to lambda (except for those entries where k was 0.) So dividing
> any one k1 entry by its corresponding non-zero k entry would
> recover lambda.
> --
> "This quitting thing, it's a hard habit to break once you start."
> -- Walter Matthau
For example, I have the [a] matrix>
a =
2 4 6 10
4 20 50 80
6 50 8 6
10 80 6 30
I need to write now the [b] matrix>
b=
8 50 6 10
4 20 30 9
7 50 8 6
10 15 6 18
in such way that L*b=a were L is a unknown constant (scalar).
Along the same line, if I have the following matriceal equation>
( [k1] + [k2] ) * {u} = {f}
[k1], [k2] - size n x n, {u} - size 1 x n and {f} - size 1 x n
I want to find L in such way that
( L*[k] ) * {u} = {f}
were only L is unknown and [k] = [k1] + [k2]
"John D'Errico" <woodchips@rochester.rr.com> wrote in message <g439hp$n2v
$1@fred.mathworks.com>...
> Why so much effort? The least squares solution
> is just this:
>
> lambda = k(:)\k1(:);
>
> John
Yes, I stand corrected, John. That gives the same answer with much less effort.
I've been making more and more of such bloopers lately, I'm afraid.
>in such way that L*b=a were L is a unknown constant (scalar).
In that case, either '*' does not represent matrix-scalar
multiplication, or L is not a scalar.
What solution have you worked out "manually" for this example?
>Along the same line, if I have the following matriceal equation>
>( [k1] + [k2] ) * {u} = {f}
> [k1], [k2] - size n x n, {u} - size 1 x n and {f} - size 1 x n
>I want to find L in such way that
>( L*[k] ) * {u} = {f}
>were only L is unknown and [k] = [k1] + [k2]
If L is a scalar, then in that situation, L = 1.
If L is a matrix, then in that situation, L is the n x n identity
matrix.
--
"We may gather out of history a policy, by the comparison and
application of other men's forepassed miseries with our own like
errors and ill deservings." -- Sir Walter Raleigh
On 27 Iun, 21:25, rober...@ibd.nrc-cnrc.gc.ca (Walter Roberson) wrote:
> In article <c285fc5e-58ac-4693-8515-87464f84c...@r66g2000hsg.googlegroups.com>,
>
>
>
> boguS <bteod...@gmail.com> wrote:
> >For example, I have the [a] matrix>
> >a =
>
> > 2 4 6 10
> > 4 20 50 80
> > 6 50 8 6
> > 10 80 6 30
>
> >I need to write now the [b] matrix>
> >b=
>
> > 8 50 6 10
> > 4 20 30 9
> > 7 50 8 6
> > 10 15 6 18
> >in such way that L*b=a were L is a unknown constant (scalar).
>
> In that case, either '*' does not represent matrix-scalar
> multiplication, or L is not a scalar.
>
> What solution have you worked out "manually" for this example?
>
> >Along the same line, if I have the following matriceal equation>
> >( [k1] + [k2] ) * {u} = {f}
> > [k1], [k2] - size n x n, {u} - size 1 x n and {f} - size 1 x n
> >I want to find L in such way that
> >( L*[k] ) * {u} = {f}
> >were only L is unknown and [k] = [k1] + [k2]
>
> If L is a scalar, then in that situation, L = 1.
> If L is a matrix, then in that situation, L is the n x n identity
> matrix.
>
> --
> "We may gather out of history a policy, by the comparison and
> application of other men's forepassed miseries with our own like
> errors and ill deservings." -- Sir Walter Raleigh
Sorry, my mistake... [k] is not equal with ( [k1] + [k2] ), [k] is
just a known matrix.
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