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From: stephen on 21 Jul 2010 09:15
Hi, I have a matrix algebra question.

I have a unkown 1x9 vector A (which is value for each security) and a known 8x9 matrix B (which has key rate DV01 for each security. And I also have another 8x1 vector C.
To solve for A,
B*A'=C
A'=inv(B)*C
A=(inv(B)*C)'

But the problem is B is not a square matrix and could not solve for its inverse. Anyone knows a way to get around it and solve for A?

Thanks for your help.

Stephen
From: John D'Errico on 21 Jul 2010 09:27
"stephen" <huangj5(a)nationwide.com> wrote in message <i26rss$l8g$1(a)fred.mathworks.com>...
> Hi, I have a matrix algebra question.
>
> I have a unkown 1x9 vector A (which is value for each security) and a known 8x9 matrix B (which has key rate DV01 for each security. And I also have another 8x1 vector C.
> To solve for A,
> B*A'=C
> A'=inv(B)*C
> A=(inv(B)*C)'
>
> But the problem is B is not a square matrix and could not solve for its inverse. Anyone knows a way to get around it and solve for A?
>
> Thanks for your help.
>
> Stephen

One solution is just

A = (B\C)';

Note that your problem is rank deficient, so one of the
unknown elements of A will be zero.

A different solution arises from pinv.

A = (pinv(B)*C)';

John
From: stephen on 21 Jul 2010 11:36
Very helpful. Thanks!


"John D'Errico" <woodchips(a)rochester.rr.com> wrote in message <i26sjo$7s3$1(a)fred.mathworks.com>...
> "stephen" <huangj5(a)nationwide.com> wrote in message <i26rss$l8g$1(a)fred.mathworks.com>...
> > Hi, I have a matrix algebra question.
> >
> > I have a unkown 1x9 vector A (which is value for each security) and a known 8x9 matrix B (which has key rate DV01 for each security. And I also have another 8x1 vector C.
> > To solve for A,
> > B*A'=C
> > A'=inv(B)*C
> > A=(inv(B)*C)'
> >
> > But the problem is B is not a square matrix and could not solve for its inverse. Anyone knows a way to get around it and solve for A?
> >
> > Thanks for your help.
> >
> > Stephen
>
> One solution is just
>
> A = (B\C)';
>
> Note that your problem is rank deficient, so one of the
> unknown elements of A will be zero.
>
> A different solution arises from pinv.
>
> A = (pinv(B)*C)';
>
> John
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