# Thread: What is the dimension of the solution set of a nonhomogeneous equation system?

1. ## What is the dimension of the solution set of a nonhomogeneous equation system?

I thought its dimension is equal to the dimension of the solution set of the same homogenous system. Am I right?

2. ## Re: What is the dimension of the solution set of a nonhomogeneous equation system?

Not necessarily. We can think of a homogeneous system of n equations as a matrix equation Ax= 0 where A is the matrix of coefficients of the system of equations. The solution set of that homogeneous system of equations is the kernel of the matrix A and its dimension is the "nullity", k, of A. by the rank-nullity property, if A maps $R^n$ to $R^m$, then th e sum of the rank and nullity of A is n. That is, A maps all of $R^n$ to a n- k dimensional subspace of $R^m$.

We can write a non-homogeneous system as Ax= b where b is the vector containing the right side of the equations. If be happens to lie in the n- k dimensional subspace that A maps all of $R^n$ into (the "image of $R^n$ under A") then the solution set has dimension k, the same as the kernel. But if b is not in that subspace there is no solution. That is what is sometimes called the "Fredholm alternative".

3. ## Re: What is the dimension of the solution set of a nonhomogeneous equation system?

often the non-homogeneous case is represented by the dictum: general solution = homogeneous solution + particular solution.

note that if there is NO particular solution, then the number of homogeneous solutions (that is, the rank of the matrix) is irrelevent.

this often happens in problems where one is to determine if b is in col(A). the smaller the rank of A, the less likely it is that this will be true,

although if it IS true, there are often several ways to combine the columns of A to get b.

(if rank(A) << n, where A is nxn, then we have multiple ways to make a basis from the columns of A).

said yet another way: just because rref(A) has rank k, does not mean the system is consistent. the augmented

matrix might have rank k+r, which would have no solutions.