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.

  • $\begingroup$ Just want to make sure I understand correctly; this statement would be false because if a zero row was present that would imply there was a non-trivial solution. Right? $\endgroup$ – Sujaan Kunalan May 30 '13 at 17:03

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.

  • $\begingroup$ Thank you! Until now, I was under the impression that a matrix had infinitely many solutions only if there was a row of zeros, or if the rank was less than the number of variables. $\endgroup$ – Sujaan Kunalan May 30 '13 at 17:27
  • $\begingroup$ This would be true if the matrix was square. Otherwise, you need to think of it in terms of free variables and counting the number of pivots in the columns. $\endgroup$ – Adriano May 30 '13 at 17:29

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.

  • $\begingroup$ So if I see a zero row in a homogenous system, I can say for certain that the solutions are non-trivial and can be given in parameters? $\endgroup$ – Sujaan Kunalan May 30 '13 at 16:54
  • $\begingroup$ if you have as many equations as you have variables (which is often the case then yes) that's true, but say I have 5 equations for only 4 variables, then clearly I must get at least one zero row but that doesn't necessarily mean I'll have infinitely many equations. $\endgroup$ – john May 30 '13 at 16:57

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..


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