# Number of linearly independent solution of a homogeneous system of equations.

The number of linearly independent solution of the homogeneous system of equations $$AX=O$$, where $$X$$ consists of $$n$$ unknowns and $$A$$ consist of $$m$$ linearly independent rows can't be equal to $$n$$

(True/False)

We know that for such a system the number of linearly independent solutions is given by $$n-m$$. When $$n-m = n$$ This implies $$m=0$$. That means number of linearly independent rows is $$0$$. So rank of $$A$$ should be $$0$$. And nullity is $$n$$. Is it true that if $$A$$ is a null matrix then there are $$n$$ linearly independent solution.

It seems like, all $$n$$ column vectors are solution for this system if $$A$$ is null matrix. And dimension of all $$n$$ column vectors is $$n$$. So the statement should be false. Am I correct$$?$$

• For such a system there is either one solution or infinitely many solutions. – user247327 Sep 17 '19 at 11:23
• Yes. You have provided an example for which the statement fails; namely, if $A$ has rank zero. – Bubaya Sep 17 '19 at 11:24
• @user247327 I know that but this problem asks about possibility of number of linearly independent solutions. – Mathaddict Sep 17 '19 at 11:38

If $$(v_1, \dots, v_n)$$ are linearly independent vectors, then $$v_i \neq 0$$ for all $$1\leqslant i \leqslant n$$.
If one of them, say $$v_1$$ is null, take $$\lambda_i = \left\{ \begin{matrix} 0 \textrm{ if } i\neq 0\\ 1 \textrm{ if } i = 1 \end{matrix} \right.$$ Then $$\sum_{i=1}^n \lambda_i v_i = 0$$ but the $$\lambda_i$$s are not identicaly $$0$$.
• @Mathsaddict Well, if $A$ is null, its column vectors are also null. I can't make sense out of the last paragraph in your question. – Olivier Roche Sep 17 '19 at 11:49
• I mean, if $A$ is a null matrix, then all n tuples in $R^{n}$ are solutions of this system. – Mathaddict Sep 17 '19 at 11:52