@Carl Love Thank you for this draft. What is your idea about adding "W[i]-1" to the set  {coeffs(expand(`+`((C*~F)[]) - f), V)} to effectively utilize the NotNull conditions? Since, the elements of W are nonzero we can set all of them equal 1. What is your idea?

@Carl Love Now I want to test linearly dependence by additional condition for linear parametric polynomial (linear for variables not parameters). 

For example let

and

(a,b,c,d,e,h are parameters and A1,A2,A3 are variables).

If I use PolyLinearCombo(F,f,{A1,A2,A3}) then its output is false,[].

Now we let to condition sets for parameters as the following:

N:=[ebc+ahd]

W:=[a,c]

The elements of N must be zero means that ebc+ahd=0

and the elements of W are non-zero that is a<>0 and c <>0.

Let a=b=c=d=h=1 and e=-1. This specialization satisfy in the above condition sets. By this specialization we have:

and

Now if I use PolyLinearCombo(F,f,{A1,A2,A3}) then its output is true,[-1,1].

By this additional two condition sets I have to check that whether f is linearly independent of F or not. How can I do this without specialization? In fact I want an algorithm that its input is (null condition, not-null condition, list of polynomials F, a polynomial f, the set of variables) and its output is true and coefficients if f is linearly dependent of F w.r.t. null and not-null conditions, else its output is false.

If the name of new procedure is ExtPolyLinearCombo and 

N:=[ebc+ahd]

W:=[a,c]

I want the output of

ExtPolyLinearCombo(N,W,F,f,{A1,A2,A3}) be true,[coefficients]

Thank you very much in advance.

@Dmitry Lyakhov 

You can use double sequences for your purpose. This is faster from loop for. See the following example:

> A := [1, 2, 3]; B := [a, b, c];
                          
> seq(seq({A[i], B[j]}, j = 1 .. nops(B)), i = 1 .. nops(A));

{1, a}, {1, b}, {1, c}, {2, a}, {2, b}, {2, c}, {3, a}, {3, b}, {3, c}

@Carl Love Yes it is good for some example so far.

@Carl Love I think it is better that in Homogenize procedure you remove V from its input and set V=[op(T)] at inside the procedure. What is your idea?

@Carl Love I need both f and F in my main procedure. It is possible that the monomials appears in F are different by monomial in f. In fact the input of this algorithm are (f , F , a monomial ordering) and its output is homogeneous ff and FF.

@MDD Consider the following example:

its output is:

Also we know that its output can be "true,[a,0,0]". This example shows that in your implementation the coefficients are not uniqe. I think this happened since the polynomials

are linearly dependent. I deduce that when the list of polynomials be linearly independent then the coefficients are uniqe. What is your idea about this example and my statements?

@Carl Love This is complete now. Thank you so much for your implementation.

@Carl Love Consider the following examples, too:

In the above examples $a$ is a nonzero parameter and the outputs must be true but both outputs are false.

@Carl Love I think that there is a minor error in your procedure, Please run the following example:

Could you please remove this error? If the polynomial is zero then it is not a problem. Please set the output true with zero coefficients.

@Carl Love This is very complete and I have to say thank you so much. I know implementation of some computer algebra algorithms but your implementation is very professional.

@Carl Love Thank you so much. I know these command! I want to know in general the way of your implementation. Thank you so much.

@Carl Love This is very good. Could you please explain your way? I dont know some command that you use in the above your implementation such as "C*" and "~" .

@Carl Love Could you please combine this way with your LinearCombo for checking whether a polynomial f is linear independent of some other polynomials F. Please note that f is a polynomial with parametric coefficients and F is a list of polynomials with parametric coefficients. I want the output is

1- true(f is linear dependent of F) and the coefficients vectors

or

2- false(f is linear independent of F)

@Carl Love Thanks for this good way. I have a suggestion: When its output is false (the polynomials are linearly independent) then the finding coefficients is meaningless. For saving memory and time it is better do not compute the coefficient, when it is false. What is your idea?