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

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
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.