# vector space proofs

• Oct 4th 2009, 02:57 PM
alexandrabel90
vector space proofs
how do you prove that if V is a vector space and v is a vector and c is any scalar, then if cv =0, it means that c=0 or v=0?

i was thinking of using the fact that since v is in V, there exist a negative vector -v in V.

so cv + (-cv) = c ( v+ (-1)v) =c0 = 0

then after that, im not sure how im going to show that c=0 or v=0..
• Oct 4th 2009, 03:12 PM
Chris L T521
Quote:

Originally Posted by alexandrabel90
how do you prove that if V is a vector space and v is a vector and c is any scalar, then if cv =0, it means that c=0 or v=0?

i was thinking of using the fact that since v is in V, there exist a negative vector -v in V.

so cv + (-cv) = c ( v+ (-1)v) =c0 = 0

then after that, im not sure how im going to show that c=0 or v=0..

Let $\displaystyle \mathbf{v}=\left<v_1,v_2,\dots,v_n\right>\in V$.

Now, $\displaystyle c\mathbf{v}=\mathbf{0}\implies c\left<v_1,v_2,\dots,v_n\right>=\left<0,0,\dots,0\ right>\implies\left<cv_1,cv_2,\dots,cv_n\right>=\l eft<0,0,\dots,0\right>$.

This implies that $\displaystyle cv_i=0,\,1\leq i\leq n$.

By the zero product property, it follows that if $\displaystyle cv_i=0$, then $\displaystyle c=0$ or $\displaystyle v_i=0\implies \mathbf{v}=\mathbf{0}$.

Does this make sense?
• Oct 4th 2009, 03:36 PM
Bruno J.
That is not bad, Chris, but you assume that the vector space has a finite basis. Some vector spaces have infinite bases, sometimes even uncountable bases, and then the argument is hard to generalize.

Suppose $\displaystyle cv=0$. If $\displaystyle c=0$, we are done. If $\displaystyle c \neq 0$, $\displaystyle c$ is invertible, and hence $\displaystyle c^{-1}cv=0 \Rightarrow v = 0$ (We have used the axiom which states that $\displaystyle a(bv)=(ab)v$ for any $\displaystyle a, b$ in the field).

Hence if $\displaystyle cv=0$, $\displaystyle v=0$ or $\displaystyle c=0$.