- #1

JD_PM

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- TL;DR Summary
- I want to prove the following: Let ##V## be a vectorspace and ##\beta## a basis for $V$. Now make a partition of ##\beta## in a disjoint union of subsets ##\beta_1, \ldots, \beta_k## and let ##U_i = \text{span}(\beta_i)## for every ##i = 1, \ldots, k##. Prove then that ##V = U_1 \oplus U_2 \oplus \ldots \oplus U_k##.

Attempt:

Take an arbitrary vector ##v \in V##. Then we have to show that there are unique vectors ##u_1 \in U_1, u_2 \in U_2, \ldots, u_k \in U_k## such that \begin{align*} v = u_1 + u_2 + \ldots + u_k. \end{align*}

We prove by contradiction: Suppose there are two such ways, i.e. that \begin{align*} v= u_1' + u_2' + \ldots + u_k'. \end{align*} also holds. Then we have \begin{align*} \sum_{i=1}^k u_i = \sum_{i=1}^k u_i', \end{align*} or \begin{align*} (u_1 - u_1') + (u_2 - u_2') + \ldots + (u_k - u_k') = 0 \end{align*} The latter equation requires that ##u_i = u_i'##, a contradiction.

Do you agree? Or is there a neater way to show it? :)

Thanks!

Take an arbitrary vector ##v \in V##. Then we have to show that there are unique vectors ##u_1 \in U_1, u_2 \in U_2, \ldots, u_k \in U_k## such that \begin{align*} v = u_1 + u_2 + \ldots + u_k. \end{align*}

We prove by contradiction: Suppose there are two such ways, i.e. that \begin{align*} v= u_1' + u_2' + \ldots + u_k'. \end{align*} also holds. Then we have \begin{align*} \sum_{i=1}^k u_i = \sum_{i=1}^k u_i', \end{align*} or \begin{align*} (u_1 - u_1') + (u_2 - u_2') + \ldots + (u_k - u_k') = 0 \end{align*} The latter equation requires that ##u_i = u_i'##, a contradiction.

Do you agree? Or is there a neater way to show it? :)

Thanks!