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When r is less than n .
I have to find a way to do this, its a question on my upcoming exam and I've been looking in a whole lotta places. I found something, I'm just not 100% it's what I need.
Basically I need to find a way to solve a linear system where r is less than n with a matrix solution that's not the Gauss method, but the matrix equation method.
Now, if I'm right about the upper equation being used for this method, what follows after it ? Do you guys know of a possible matrix solution of linear systems that starts out like that ?
I found one way in a book, it's basically used when det A == 0 . It's a formula that goes like this:
The A and X in the above formula have the character ~ above them.
I actually misunderstood the question, it seems it was just a general question about solving a system when r is less than n. I'm still curious about this, though, I've seen it once on some lectures but I didn't write it down. And I can't seem to remember if it has anything to do with the formula I wrote above
I might also point out that, in addition to saying "r is less than n", you ought to tell us what "r" and "n" mean! I suspect that one is the number of equations (number of rows of the matrix) and the other is the number of unknowns (number of columns of the matrix) but i have to guess that and, in any case, I don't know which is which.
Yeah, my bad. I guess there isn't a particular standard for all mathematicians as far as the use of these letters (I know it's different in different fields, but I'm talking about the same field - matrices) ?
R is the rank of the system matrix, and N is the number of unknowns.
If r< n- that is if the rank is less than the number of rows, you cannot use this because A is not invertible- does not exist.
Does the book say what , , , and mean?I found one way in a book, it's basically used when det A == 0 . It's a formula that goes like this:
The A and X in the above formula have the character ~ above them.
You could do this: If A: does not have an inverse then it maps all of into a sub space of . There is no solution to Ax= B unless B happens to lie in that subspace (in which case there may be an infinite number of solutions). If B is not in the subspace then the best you can do is find x so that Ax is "closest" to B. To do that, drop a perpendicular from B to the subspace. Call vector, the vector in the range of A closest to B, B' and look at B- B'= B- Ax where x is such that Ax= B'. That vector is perpendicular to the subpace: <B- Ax, u>= 0 for all u in the subspace. But every vector in that subspace is of the form u= Av for some v because that subspace is the range of A, that is the same as <B- Ax, Av>= 0. Now define A~ to be the "adjoint" of A, the linear operator such that <v, Au>= <A*v, u> for all u in and all v in (for real matrices this is the transpose).
Then we have <B- Ax, Av>= <A*(B- Ax), v> where this inner product is on and v can be any vector in / In particular, v couls be A*(B- Ax) itself so we must have A*(B- Ax)= A*B- A*Ax= 0 or A*Ax= A*B. Now it may happen that A*A is invertible even if A itself is not. In that case we have [itex]x= (A*A)^{-1}A*B[/itex], and [itex](A*A)^{-1}A*[/itex] is called the "generalized inverse". If A itself has an inverse, then [itex](A*A)^{-1}= A^{-1}A*^{-1}[/itex] so that [itex](A*A)^{-1}A*= A^{-1}A*^{-1}A*= A^{-1}[/itex] the true inverse.
Note that * throughout means the "adjoint", not multiplication.
Damn, I didn't understand that post well. The way we are taught matrices here is pretty simplistic, I guess. I don't think this stuff was even applied in space like you're doing it in the first part of your post.
Anyway, let me try explain that formula using an example.
is a vector that contains the parameters that multiply with , so in this case 3 and -3.
is a vector that contains the unknowns that determine the other unknowns (in this case, it's only .
is a 2x2 matrix with the multipliers of the other unknowns (1, -1, 2 and 1)
is another vector, with the numbers on the right side of the mentioned equation (1 and 1)
The 'r' in the subscript means reduced, and as I understood it, is the reduced matrix of A in the AX=B equation. So is the highest ranking matrix for which the determinant is different from 0.
Am I making any sense here ? :P