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

[JD_PM]

TL;DR

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!

 

[fresh_42]

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 $1\leq i \neq j\leq k.$

 

[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 $U_i\cap U_j=\{0\}$

 

[PeroK]

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.

 

[PeroK]

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 $x \in B$.

 

[JD_PM]

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.

 

[PeroK]

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 $a = b = 0$.