# Linear Algebra Row Space solution

Let's suppose we have a matrix A mxn that implements a linear system $Ax=b$.
If b is inside the Column Space of A, then thel linear system has AT LEAST one solution x in $R^n$.
But every x in $R^n$ can be represented by a sum of two vectors, each one inside of a ortogonal complement space of $R^n$.
Let's suppose we take N(A) and $C(A^T)$.They are clearly ortogonal complements subspaces of $R^n$.
So, if x is a solution $R^n$, it can be represented as $x = r + n$ , where r is a vector inside the rowspace and n is a vector inside the NullSpace.

Treating x = r+n like a solution, it would satisfy Ax = b :
A(r+n) = Ar + An = Ar = b

That means for any solution x, it can be decomposed in x = r + n, and we can find an even smaller solution r, represented by a vector that lies in the rowspace of A.

Also, there are kind of solutions to Ax =b :
1 - Unique Solution
2 - Infinitely many solutions
3 - No Solutions
My question is if this holds only for the second case ( if Ax=b has infinity solutions ) because N(A) is non-trivial.

Thanks

• You seem to understand quite much about this theory. But.. what is your question exactly? In this generality it is not sure that $N(A)$ is nontrivial. Mar 27, 2013 at 21:27
• ? Can you not have infinitenly many solutions with N(A) non-trivial ? Mar 27, 2013 at 21:52
• No; if $\mathbf{0} \ne \mathbf{n} \in N(A)$, then $\mathbf{x+n}$ is also a solution, so you must have infinitely many solutions. Of course, if there are no solutions, then $\mathbf{x+n}$ will not be a solution. Mar 27, 2013 at 22:07

In the first case, there is a unique solution to $A\mathbf{x}=\mathbf{b}$, so $N(A)=\{\mathbf{0}\}$ and we can write $\mathbf{x}=\mathbf{x}+\mathbf{0}$, where $\mathbf{x} \in C(A^T)=\mathbb{R^n}$ and $\mathbf{0} \in N(A)$. In the third case, there are no solutions to $A\mathbf{x}=\mathbf{b}$, so any statement about $\mathbf{x}$ is vacuously true. Hence the statement holds for all cases.