Let $\textbf{u}$ is any vectors, and $c$ is any scalars, prove that if $c \textbf{u} = \textbf{0}$ then either $c = 0$ or $\textbf{u} = \textbf{0}$.

Question

I had written the statement as shown.

Using contrapostion, let $c \neq 0$ and $\textbf{u} \neq \textbf{0}$. Therefore, it is obvious that $c \textbf{u} \neq \textbf{0}$.

I can't find the formal way to prove it. Is there anybody who can give me a hint or a solution?

Answer 1

If $c=0$ we are done. Suppose $c\neq 0$. Then operating $c^{-1}$ on both the sides of the given expression we get $c^{-1}c u=c^{-1}0\implies 1 u=0\implies u=0.$

Answer 2

Multiplying a scalar (here $c$) to a vector (here $u$), scales the $u$ vector, by $c$ times in magnitude and there is no change in its direction.

So, $cu = 0$ means scaling the $u$ vector by a scalar made it to be a zero vector, which has zero magnitude. Keep in mind of the concept in arithmetics, that "If we obtain zero on multiplying two numbers, then one of them is zero". Thus, there are only two possibilites in our case.

1. Either the vector $u$ itself is a zero vector, which is nothing but $u = 0$.
2. Or, the scalar we multiplied with ($c$), scaled the vector zero times, thus leading to a zero vector. This is nothing but $c = 0$.

Hope this helps.