Timeline for Check whether an overdetermined linear equation system is consistent: general approach?
Current License: CC BY-SA 2.5
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 18, 2010 at 9:18 | vote | accept | Graviton | ||
| Dec 16, 2010 at 7:16 | comment | added | mpiktas | Matrix $A$ dimension is $n\times k$, matrix $Q$ dimension is $n\times n$, matrix $R$ dimension is $n\times k$. Vector $x$ dimension is $k\times 1$, vector $b$ dimension is $n\times 1$, so $Rx$ dimension is $n\times 1$, and $Q^Tb$ dimension is $n\times 1$. Everything is ok. | |
| Dec 16, 2010 at 6:40 | comment | added | Graviton | @mpiktas, I think your above formulation is still wrong. Take this equation: $Rx=Q^Tb$. Note that the LHS the dimension is $n \times 1$, but the RHS the dimension is $k \times 1$. There is a dimension mismatch here. | |
| Dec 15, 2010 at 11:43 | comment | added | mpiktas | @Ngu Soon Hui. When we get to the matrix $R_1$ and vector $b_1$ we get the square linear system. Since determinant of $R_1$ is non zero, the solution always exists. If $Q^Tb$ is not of the form I mentioned, then the solution of the original system does not exist. | |
| Dec 15, 2010 at 11:41 | history | edited | mpiktas | CC BY-SA 2.5 | added the explanation how to calculate the solution |
| Dec 15, 2010 at 11:40 | comment | added | Graviton | @mpiktas, that's the problem: without $x$, how to check whether the equation $Rx=Q^Tb$ is fulfilled? | |
| Dec 15, 2010 at 11:34 | comment | added | mpiktas | @Ngu Soon Hui, you do not have to if you only want to check whether $x$ is the solution. | |
| Dec 15, 2010 at 11:22 | comment | added | Graviton | @mpiktas, even with your fix, I still fail to see how to compute $x$. | |
| Dec 15, 2010 at 9:36 | comment | added | mpiktas | OK, I fixed the answer. Actually there is a slight typo in the question, $b$ should be $n\times 1$ vector, not $k\times 1$. | |
| Dec 15, 2010 at 9:33 | history | edited | mpiktas | CC BY-SA 2.5 | added 460 characters in body |
| Dec 15, 2010 at 9:17 | comment | added | mpiktas | @Ngu Soon Hui, sorry my answer is incorrect. I will see if I can fix it. | |
| Dec 15, 2010 at 9:16 | comment | added | mpiktas | Yes, you are right. You cannot even multiply $Q$ by $b$, since the dimensions do not match. The answer I think still lies in decomposition of $A$, but it should be rephrased better. | |
| Dec 15, 2010 at 9:16 | comment | added | Graviton | @mpiktas, how to compute the $x$ in this case? | |
| Dec 15, 2010 at 9:03 | comment | added | J. M. ain't a mathematician | mpiktas, that's for least squares. For instance: if you QR decompose the 3-by-2 example of Ngu (call the original matrix $\mathbf A$), and let $\mathbf b=(2\quad 3\quad 4)^T$ and $\mathbf x=\left(0\quad \frac1{10}\right)^T$ , your equation doesn't work, even if the $\mathbf x$ I gave minimizes $\|\mathbf A\mathbf x-\mathbf b\|_{\infty}$ | |
| Dec 15, 2010 at 8:09 | history | answered | mpiktas | CC BY-SA 2.5 |