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Is there any particular matrix $(B)$ such that after multiplying to orginal matrix ($A_{m*n}$) gathering information of the matrix on diagonal of resulting matrix $(C)$??

in other words, "dependent columns" of $A$ after multiplying to $B$ would result zero elements on diagonal of $C$ and independent columns result non-zero elements.

Is it possible??

ex. $$A=\begin{pmatrix} a & b & c \\ d & e & f \\ g& h&i\end{pmatrix}$$ where first and second column are basis vectors of A (rank is 2) . I'd like define $B_{n*n}$ such that $$B=\begin{pmatrix} a' & b' & c' \\ d'& e' & f' \\ g'& h'&i'\end{pmatrix}$$ then $$AB=\begin{pmatrix} a'' & b'' & c'' \\ d''& e'' & f'' \\ g''& h''& 0\end{pmatrix}$$ $a'',e''$ are not $0$

hoom
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    It's very hard to understand exactly what you're asking. Can you write some equations to express your goal? Can you give a hypothetical example? – dfeuer Dec 18 '13 at 15:44
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    Some ways you can make your question more understandable: use the term "multiplying" rather than "producting" (which is not a word). Use equations to explain what you want. It is not at all clear what you mean by "dependent columns of $A$ after producting to $B$ would result zero elements on diagonal of $C$". What are "dependent columns?" Is $C$ simply given by $C = AB$? Or do you mean $C = BA$? I have no idea what you could possibly mean by "gathering information". – Ben Grossmann Dec 18 '13 at 16:24
  • @Sarvenaz Have another look at my answer, which I have just re-written to address what seems to be what you're aiming at, per your added part of the posted question. – coffeemath Dec 18 '13 at 19:51

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If your matrix $A$ has its first two columns independent, and the third column dependent on the first two columns (so $A$ has rank 2), then you can work with the transpose $A^t$ which will then have its last row dependent on its first two rows, where the first two rows are independent. Row reduction will then put $A^t$ into the shape $$C=\pmatrix{1&0&x\\0&1&y\\0&0&0}$$ and to get the matrix $Q$ which when multiplying $A^t$ on the left gives $C$ one technique is to start with a 3 by 6 matrix consisting of $A^t$ followed by a 3 by 3 identity matrix. Then after row reduction (only carried out on the left 3 by 3 part, viewing the 3 by 3 section after that as "right hand sides") the desired matrix $Q$ will be located where the identity matrix was initially. Then go by transpose again and your original matrix $A$ should be postmultiplied by the transposed matrix $Q^t.$

Example: Original matrix $$A=\pmatrix{1&2&3\\2&3&5\\3&4&7}.$$ Taking transpose and inserting 3 by 3 identity afterwards gives $$\pmatrix{1&2&3&1&0&0\\2&3&4&0&1&0\\3&5&7&0&0&1}.$$ Row reduction of this (only on the left 3 by 3 part) gives $$\pmatrix{1&0&-1&-3&2&0\\0&1&2&2&-1&0\\0&0&0&1&1&-1}.$$ Then $Q$ is read off as the second 3 by 3 part of this, namely

$$Q=\pmatrix{-3&2&0\\2&-1&0\\1&1&-1}.$$ Then $QA^t$ is equal to the left 3 by 3 matrix of the reduced matrix above, namely $$QA^t=\pmatrix{1&0&-1\\0&1&2\\0&0&0}.$$ Finally one takes transposes to get $AQ^t$ as the matrix you seek, in this case $$C=\pmatrix{1&0&0\\0&1&0\\-1&2&0}.$$ In general for the rank 2 situation you describe, with the first two columns independent, the final matrix $c$ obtained this way will have $1$ on the first two diagonal elements (so nonzero as you request) and also necessarily a $0$ in the lower right diagonal place. There are also zeros in positions $(1,2),(2,1),(1,3),(2,3)$ if one has gone to complete row reduction at the corresponding stage of row reducing the transpose of $A$.

coffeemath
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  • Thanks to whoever edited this for introducing me to "pmatrix" ... – coffeemath Dec 18 '13 at 16:09
  • Thanks, perhaps I did not explain correctly. zero elements for dependent columns of A but what is independent columns of A?? the corresponding elements in C matrix must be non-zero if such matrix (C) exists – hoom Dec 18 '13 at 16:16
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    @coffeemath: I'll do you one better. When using \pmatrix in MathJax, you don't need the \begin and \end statements; you could just do something like $\pmatrix{a&b\\c&d}$. – Ben Grossmann Dec 18 '13 at 16:18
  • @Sarvenaz In your post you say you want the diagonal of $C$ to be zero. Your statement about dependent columns of $A$ is hard to decipher. It would be good if in your question you could give a concrete example of what the question is you're asking. – coffeemath Dec 18 '13 at 16:21
  • @coffeemath: another question :) Is there any less computational way that works well in high-dimensional matrices? Thanks. – hoom Dec 18 '13 at 21:04
  • @Sarvenaz I don't know of a faster way. Basically one is dealing with keeping track of row reduction via elementary matrices, and the "trick" of appending an identity matrix while carrying out the row reduction seems to get the required product of elementary matrices without finding each separately, so is itself somewhat of a short-cut. But since the starting matrix might be anything, I doubt if one could get the required multiplier directly in terms of the entries of the starting matrix. – coffeemath Dec 18 '13 at 22:00
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    That was my edit switching matrix to pmatrix. @Omnomnomnom, it looks like the begin/end pmatrix is The $\LaTeX$ Way, while \pmatrix is Plain $\TeX$. No, I do not understand why $\LaTeX$ prefers any solution that involves more typing. – dfeuer Dec 19 '13 at 03:25
  • @dfeuer I had been wondering why \pmatrix would work with MathJax and not with $\LaTeX$, thanks for shedding a little light on that. What's even more confusing is that \bmatrix is not a valid command. No idea why $\LaTeX$ would come to that solution, maybe it's worth asking at the TeX SE. – Ben Grossmann Dec 19 '13 at 04:12
  • @Omnomnomnom, I suspect Knuth considered bmatrix a cheap substitute for pmatrix for inferior typesetting systems. – dfeuer Dec 19 '13 at 05:59
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    @dfeuer: pmatrix is not a TeX primitive. It has nothing to do with Knuth. pmatrix was introduced in amstex, so follows the plain TeX convention for stuff. With the more modern amsmath packages it has been superceded by the \begin{...} ... \end{...} construction favored by LaTeX. It involves more typing but it is much, much easier to read: your eyes have an easier time finding \end{pmatrix} than just the (correct) closing brace. – Willie Wong Dec 19 '13 at 09:01
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    And in plain TeX the more correct construction for environments should be \pmatrix ... \endpmatrix, see for example, this TeX.SE question. The choice to use the \pmatrix{...} syntax and the choice to not include an implementation for \bmatrix{...} is entirely due to the MathJax team, and has nothing to do with TeX versus LaTeX. But this is getting really off topic. – Willie Wong Dec 19 '13 at 09:03