MHB Tyler's question at Yahoo Answers (linear independence)

[Fernando Revilla]

Here is the question:

Show that if {w1, ... wm} is linearly independent in V, then {[w1]B , ... [wm]B} is linearly independent in Rn.

Here is a link to the question:

Linear algebra help please please? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[Fernando Revilla]

Hello Tyler,

Fixing a basis $B=\{w_1,\ldots,w_m\}$ of a real vector space $V$ we know that

$(i)\;[x+y]_B=[x]_B+[y]_B$ or all $x,y\in V$.
$(ii)\;[\lambda x]_B=\lambda [x]_B$ for all $\lambda \in \mathbb{R}$.
$(iii)\;[x]_B=(0,\ldots ,0)\Leftrightarrow x=0$.

Suppose $\lambda_1 [w_1]_B +\ldots +\lambda_m [w_m]_B=(0,\ldots ,0)$. Using $(i)$ and $(ii)$:
$$[\lambda_1 w_1 +\ldots +\lambda_m w_m]_B=(0,\ldots, 0)$$
Using $(iii)$: $\lambda_1 w_1 +\ldots +\lambda_m w_m=0$.

By hypothesis $ \{w_1,\ldots ,w_m\}$ are linearly independent, so $\lambda_1=\ldots =\lambda_m=0$. This proves that $\{[w_1]_B,\ldots,[w_m]_B\}$ are linearly independent.