# Proving that ##V = U_1 \oplus U_2 \oplus \ldots \oplus U_k##

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• JD_PM

#### JD_PM

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!

No, not really. We have to show that ##U_1 +\ldots+ U_k \supseteq V## which is immediately clear by writing
##v=u_1+u_2+…+u_k.##
You should have been more detailed here. Set ##\beta_{i}:=\{u_{i1},\ldots,u_{in_i}\}.## Then we have ##v=\sum_{i=1}^k\sum_{j=1}^{n_i}c_{ij}u_{ij}##. Now define ##u_i:=\sum_{j=1}^{n_i}c_{ij}u_{ij}## for all ## i ##. Then
##v=u_1+u_2+…+u_k.##
##\in U_1+\ldots+U_k## and thus ##V\subseteq U_1+\ldots+U_k.##

This would have been better than merely writing
##v=u_1+u_2+…+u_k.##
which should have been the conclusion, not the first line.

Finally, you have to show now, that ##U_i\cap U_j=\{0\}## for all ##1\leq i \neq j\leq k.##

Last edited:
JD_PM
I appreciate the detailed explanation!

Given that we want to prove an equality, shouldn't we also discuss why the inclusion ##U_1 +\ldots+ U_k \subseteq V## holds?

My reasoning: the sum of subspaces of ##V## yields a subspace of ##V## (proof).

Finally, you have to show now, that ##U_i\cap U_j=\{0\}## for all ##1\leq i \neq j\leq k.##

Here's my attempt:

Let ##x \in U_i\cap U_j##. Then ##x = \underbrace{x}_{\in U_i} + \underbrace{0}_{\in U_j}## and ##x = \underbrace{0}_{\in U_i} + \underbrace{x}_{\in U_j}##.

Given that ##\beta_{i}:=\{u_{i1},\ldots,u_{in_i}\}## are disjoint subsets, the only element subspaces ##U_i## and ##U_j## have in common is the zero element. Hence, it follows that ##x=0## and ##U_i\cap U_j=\{0\}##

I appreciate the detailed explanation!

Given that we want to prove an equality, shouldn't we also discuss why the inclusion ##U_1 +\ldots+ U_k \subseteq V## holds?

My reasoning: the sum of subspaces of ##V## yields a subspace of ##V## (proof).

Here's my attempt:

Let ##x \in U_i\cap U_j##. Then ##x = \underbrace{x}_{\in U_i} + \underbrace{0}_{\in U_j}## and ##x = \underbrace{0}_{\in U_i} + \underbrace{x}_{\in U_j}##.

Given that ##\beta_{i}:=\{u_{i1},\ldots,u_{in_i}\}## are disjoint subsets, the only element subspaces ##U_i## and ##U_j## have in common is the zero element. Hence, it follows that ##x=0## and ##U_i\cap U_j=\{0\}##
This is not right. You need to use the linear independence of the basis vectors.

One of your problems is poor technique. For example, whenever you have ##x \in A \cap B## the next line automatically should be ##x \in A## and ##x \in B##.

This is not right. You need to use the linear independence of the basis vectors.

OK, might you please provide more details?

One of your problems is poor technique.

I take this comment as constructive criticism. I do my best to get better.

For example, whenever you have ##x \in A \cap B## the next line automatically should be ##x \in A## and ##x \in B##.

Alright.

Here's my attempt:

Let ##x \in U_i\cap U_j##. Then ##x = \underbrace{x}_{\in U_i} + \underbrace{0}_{\in U_j}## and ##x = \underbrace{0}_{\in U_i} + \underbrace{x}_{\in U_j}##.
It's difficult to comment on that because it doesn't look like mathematics. It's like you are trying to formulate your ideas without the proper technique.

To follow on from the above:
$$x \in U_i \ \Rightarrow \ x = a\beta_i$$ for some scalar ##a##. Then the same argument for ##j## leads to ##a\beta_i = b\beta_j## which contradicts linear independence, unless ##a = b = 0##.

JD_PM