# A $3\times3$ matrix with rank 2 must have a nonzero solution to $Ax=0$

I have recently come across the following statement:

A $3\times3$ matrix A with rank $2$ must have a nonzero solution to $Ax=0.$

I am having trouble understanding why there must be a nonzero solution to $Ax = 0.$ Is it due to the fact that we will have a free variable?

• Yes, that's one way to see it: we will have a free variable, which means we will have a non-zero solution. – Omnomnomnom Dec 20 '16 at 17:46

If the rank of a matrix is 2 then, by the Rank-Nullity theorem, the kernel must have dimension 1.

• Just for clarification: If the kernel has dimension 1, that implies that $Ax = 0$ has infinitely many solutions dependent on one variable? – h94 Dec 20 '16 at 18:06
• As @Oiler mentioned in another answer, this technically depends on the field. If I assume that you are working over the real or complex numbers, then yes, there are infinitely many solutions which can be parametrized by one parameter: let $v\in V$ be a non-zero solution to $Av =0$. Then $\{\alpha v\colon \alpha \in \mathbb{R} (\mbox{ or } \mathbb{C})\} = \mathrm{ker} A$, and so the null space is parametrized by $\alpha$. (Note that this $\alpha$ corresponds precisely to the "free parameter" you would get from row reducing, etc.) – Matt Dec 20 '16 at 18:17

Yes. If you inspect the row echelon form, you get a zero row and hence there is $1$ free parameter.

In fact, you obtain infinitly many non-zero solutions.

• Infinitely many non-zero solutions is true only if you are working over an infinite field. – Oiler Dec 20 '16 at 18:01
• @Oliver, I agree, I was assuming $\mathbb{C}$ just now. – Siong Thye Goh Dec 20 '16 at 19:31

What is your definition of rank? If rank of matrix is defined as maximal number of it's linearly independent columns, from this we immediately get that columns $c_1,c_2,c_3$ of matrix $A$ must be linearly dependent whenever $\operatorname{rank}(A)<3$. This means that $$\lambda_1 c_1 + \lambda_2 c_2 + \lambda_3 c_3 = A \left[ \begin{matrix} \lambda_1 \\ \lambda_2 \\ \lambda_3 \end{matrix} \right] = 0$$ for some $\lambda_1,\lambda_2,\lambda_3 \neq 0$.