question on matrix operations

[completelyuseless]

Hi all,

my question regards a row matrix and vector matrix specifically.

if i have a vector and a row and multiply them to get one scalar value, is it possible at all to get the vector from the row and the scalar? (or the other way around, get the row from the vector and the scalar?)

the only constraints that i know of is that either the vector or the row can only be positive values and can contain zero value elements. also this is to digitally synthesized. this must be a general method as well.

just wanted to know if this is mathematically possible, ie changing the subject of the formula.

thanks in advance,
kind regards

 

[zorro]

Hi,

The product of a row vector A by a column vector B is a scalar C:
[a1 a2 ... aN]*[b1 b2 ... bN]' = C
(' stands for transpose).
Given A and C, can B be determined?

Observe that there are many solutions with only one nonzero element, i.e. (assuming a1 and a2 are not zero):
B = [C/a1 0 0 ... 0]' is a solution
B = [0 C/a2 0 ... 0]' is another solution
...etc...

In general, assuming ai are non zero:

B = [k1/a1 k2/a2 ... kN/aN]'

such that k1+k2+...+kN=C
is a solution.
If there are zero elements ai=0, the corresponding bi can be any number.

Regards

Z

 

[_Eduardo_]

No way. Unless you put extra conditions.

Given an arbitrary row matrix P and a scalar k = (c - A*P')/(A*A')

Then any B = P' + k·A' is solution of A*B = c


(matrix: uppercase , scalar: lowercase)

 

[completelyuseless]

thanks for the help,

an alternative solution to my problem may concern matrix rather than vectors and rows.

If i had a vector of say 16 elements. (natural integers ranging from 0 to 1023), could there exist a square matrix that when multiplied always gives me a row that has identical elements.
its maybe a bit too wishful in hoping that thered be one perfect matrix for this. but is it possible to have a set of matrices that work in ranges of possible input vectors? or will each possible vector have a unique matrix?

After this, with that row of identical elements, develop the linear equations that can solve back to the original vector? Or am i missing the plot completely?

sorry if these questions are irrelevant to this board.

thanks again

kind regards

 

[_Eduardo_]

The equation A * B = c with A and c given has infinite solutions.

Depending on the type of problem may be sufficient a particular solution (the simplest is B = (c / (A * A ')) * A') but others need the solution or the family of solutions meet certain conditions .
If you need a unique solution, then it must impose conditions. But if it does not matter anyone in particular, use the above expression.


Regarding square matrices multiplied by a vector be a vector with all elements equal.
Worse still, each column of the matrix has infinite solutions independent of the other columns.