1
$\begingroup$

If C is 3 x 2 matrix, D is 2 x 3 matrix, and E is 3 x 3 matrix, that K should be 2 x 2 matrix if CKD = E. With specific entries of each matrices, I could get the elements of K with expanding the equation with matrix multiplication. But, is there another way to determine K that briefer than to expanding that equation? Because matrix C and D can't be inverted due to the different numbers of each rows and columns.

$\endgroup$
  • $\begingroup$ You could use the pseudoinverses $(C^+,D^+)$ of the matrices $(C,D)$ to write $$K=C^+ED^+$$Of course, then you'll need to figure out how to calculate those pseudoinverses. $\endgroup$ – greg Oct 21 '18 at 16:56
1
$\begingroup$

Delete the third rows of C and E, and the third columns of D and E.
Perhaps a different row and column instead if necessary.

$\endgroup$
  • $\begingroup$ whoaa, it leads to the same answer if solving it with expanding the equation. But, why we have to delete the third rows of C and E and third columns of D and E? Is there any specific rules for doing this elimination? $\endgroup$ – Dziban N Oct 21 '18 at 9:40
  • $\begingroup$ The third row is extra information that you don't need to solve the equation. If the first two rows of C were linearly dependent, you would delete one of those, and the corresponding row of E. $\endgroup$ – Empy2 Oct 21 '18 at 9:43
  • $\begingroup$ how can we know that the third row of C and the third column of D are an extra information of the equation? $\endgroup$ – Dziban N Oct 21 '18 at 10:14
0
$\begingroup$

Multiply both sides by $C'$ (from the left), where the first row of $C'$ is orthogonal to the second column of $C$, and the second row of $C'$ is orthogonal to the first column of $C$. Such vectors exist and are easy to find.
Multiply both sides (from the right) by $D'$ whose first column is orthogonal to the second row of $D$, and vice versa.
We get $C'CKDD'=C'ED',$ where $C'C$ and $DD'$ are $2×2$ diagonal matrices. It is quick to finish.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.