Non-trivial solutions implies row of zeros? If there exist non trivial solutions, the row echelon matrix of homogenous augmented matrix A has a row of zeros.
True or False?
I'm not sure where to begin as to see why this would be true or false. I know that if there are a row of zeros it means that there are infinitely many solutions, but not sure how I can tell if that means there are non-trivial solutions though. Any help would be appreciated. Thanks in advance.
 A: Recall that a system can have either $0$, $1$, or infinitely many solutions. Thus, the fact that there is at least one nontrivial solution (other than the trivial solution consisting of the zero vector) implies that there are infinitely many solutions. Thus, your statement is false; as a counterexample, consider the folloring homogeneous augmented matrix (conveniently in reduced row echelon form):
$$ A=
\left[ \begin{array}{ccc|c}
1 & 0 & 2 & 0 \\
0 & 1 & 3 & 0
  \end{array}\right]
$$
Notice that $A$ has infinitely many solutions (the third column has no pivot, so the system has one free variable), yet there is no row of zeroes.

Note: The converse is not necessarily true either. That is, it is NOT the case that: if the row echelon matrix of a homogenous augmented matrix A has a row of zeroes, then there exists a nontrivial solution. As a counterexample, consider:
$$ A=
\left[ \begin{array}{cc|c}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 0
  \end{array}\right]
$$
Notice that $A$ has only the trivial solution (every column has a pivot, so the system has no free variables), yet $A$ has a row of zeroes.
A: The trivial solution is the unique solution where every variable equals zero hence if you have infinitely many solutions, you have non-trivial solutions.
However, one thing you'll need to be careful of is how many equations you have and how many variables you have.
A: By using elementary row operations to change a row (say the bottom row of a matrix $M$) into zeros, you have shown that the bottom row is a nonzero combination of the other rows. 
By backtracking and figuring out the coefficients and putting them in a vector $\vec{x}$, it shows that $\vec{x}M=0$, where $\vec{x}\neq \vec{0}$. That $\vec{x}$ is a nontrivial solution to the matrix equation..
