Show that ##U_1 + U_2 + ... + U_k = span (U_1 \cup U_2 \cup ... \cup U_k)##

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Summary:
I want to prove that ##U_1 + U_2 + ... + U_k = span (U_1 \cup U_2 \cup ... \cup U_k)## by induction
1) Base case ##k=2##

##U_1 + U_2 = span (U_1 \cup U_2)##, which I understand how to prove is OK.

2) Induction hypothesis

We assume that the following statement holds

$$U_1 + U_2 + ... + U_{k-1} = span (U_1 \cup U_2 \cup \ ... \ \cup U_{k-1})$$

3) Induction step

$$U_1 + U_2 + ... + U_k = \underbrace{U_1 + U_2 + ... + U_{k-1}}_{=span (U_1 \cup U_2 \cup \ ... \ \cup U_{k-1})} + U_k = \ ? $$

But my issue is that I do not see how to move from ##span (U_1 \cup U_2 \cup \ ... \ \cup U_{k-1})+U_k##

Might you guide me on how to proceed further?

Thanks! :biggrin:
 

Answers and Replies

  • #2
PeroK
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To prove two sets, ##A## and ##B##, are equal you often show that ##x \in A## implies ##x \in B## and vice versa.
 
  • #3
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Hi @PeroK !

To prove two sets, ##A## and ##B##, are equal you often show that ##x \in A## implies ##x \in B## and vice versa.

Oh, I see! So let's forget about induction.

We take ##x=u_1 + u_2 + \dots + u_k \in U_1 + U_2 + \dots + U_k## where ##u_i \in U_i##.

Given that each element ##u_i## is found exclusively in ##U_i##, there is no difference between ##U_1 + U_2 + \dots + U_k## and ##U_1 \cup U_2 \cup \dots \cup U_k##. (let me know if you disagree with this statement).

Taking a linear combination of the elements ##u_1 + u_2 + \dots + u_k## leads to ##x = \sum\limits_{j = 1}^k \beta_j u_j \in span (U_1 \cup U_2 \cup \dots \cup U_k)##

Let's now go the other way and take ## x = \sum\limits_{j = 1}^r \beta_j u_j \in span (U_1 \cup U_2 \cup \dots \cup U_k)##. Next we construct sets: let ##K_1## be the set of indices ##j## such that ##u_j \in U_1##, ##K_2## be the set of indices ##j## not in ##K_1## such that ##u_j \in U_2## and so on up to ##u_j \in U_k##

Given that ##K_i## sets are disjoint, ##x = \sum\limits_{i = 1}^n \sum\limits_{j \in K_j} \beta_j u_j## and ##\sum\limits_{j \in K_i} \beta_j u_j \in U_i##, we get ##x \in U_1 + U_2 + \dots + U_k##

How do you see it? :)

PS: For what kind of proofs is best to use induction though?
 
  • #4
PeroK
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Hi @PeroK !



Oh, I see! So let's forget about induction.
That approach could apply to the inductive step!
 
  • #5
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OK I misunderstood you then. Do you have any comments on #3 though? Or you want me to focus only on induction.
 
  • #6
PeroK
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OK I misunderstood you then. Do you have any comments on #3 though? Or you want me to focus only on induction.
Hi @PeroK !



Oh, I see! So let's forget about induction.

We take ##x=u_1 + u_2 + \dots + u_k \in U_1 + U_2 + \dots + U_k## where ##u_i \in U_i##.

Given that each element ##u_i## is found exclusively in ##U_i##, there is no difference between ##U_1 + U_2 + \dots + U_k## and ##U_1 \cup U_2 \cup \dots \cup U_k##. (let me know if you disagree with this statement).

Taking a linear combination of the elements ##u_1 + u_2 + \dots + u_k## leads to ##x = \sum\limits_{j = 1}^k \beta_j u_j \in span (U_1 \cup U_2 \cup \dots \cup U_k)##

Let's now go the other way and take ## x = \sum\limits_{j = 1}^r \beta_j u_j \in span (U_1 \cup U_2 \cup \dots \cup U_k)##. Next we construct sets: let ##K_1## be the set of indices ##j## such that ##u_j \in U_1##, ##K_2## be the set of indices ##j## not in ##K_1## such that ##u_j \in U_2## and so on up to ##u_j \in U_k##

Given that ##K_i## sets are disjoint, ##x = \sum\limits_{i = 1}^n \sum\limits_{j \in K_j} \beta_j u_j## and ##\sum\limits_{j \in K_i} \beta_j u_j \in U_i##, we get ##x \in U_1 + U_2 + \dots + U_k##

How do you see it? :)

PS: For what kind of proofs is best to use induction though?
Are you sure that the ##U_i## are disjoint? Are they not just any subspaces?

What I might have tried to do is to take ##x \in span(U_1 \cup \dots U_{k+1})## and show that ##x = u + u_{k+1}##, where ##u \in span(U_1 \cup \dots U_{k})## and ##u_{k+1} \in U_{k+1}##.

Then the inductive step should follow. Induction should simplify things, although it's the same basic idea.
 
  • #7
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Are you sure that the ##U_i## are disjoint? Are they not just any subspaces?

A more detailed proof without using induction can be found here.
 
  • #8
PeroK
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A more detailed proof without using induction can be found here.
I still think that using induction simplifies things. In particular, all we really need to show is that:
$$span(U_1 \cup \dots \cup U_{k+1}) = span(U_1 \cup \dots \cup U_k) + U_{k+1} $$
The RHS is clearly a subset of the LHS, so we need to show that:
$$v \in span(U_1 \cup \dots \cup U_{k+1}) \ \Rightarrow \ v \in span(U_1 \cup \dots \cup U_k) + U_{k+1}$$
An outline of the proof is that for such a vector ##v##, we take all the terms that involve vectors in ##U_{k+1}## and (using the fact that ##U_{k+1}## is a subspace) recognise that this gives a single element of ##U_{k+1}##. We also recognise that what is left is an element of ##span(U_1 \cup \dots \cup U_k)##.

You can formalise this easily. And this should be simpler than having all those different sets of indices.
 
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  • #9
StoneTemplePython
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edit: better to just echo what PeroK said.
 
Last edited:
  • #10
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Thanks for the clarification @PeroK !

The RHS is clearly a subset of the LHS, so we need to show that:
$$v \in span(U_1 \cup \dots \cup U_{k+1}) \ \Rightarrow \ v \in span(U_1 \cup \dots \cup U_k) + U_{k+1}$$
An outline of the proof is that for such a vector ##v##, we take all the terms that involve vectors in ##U_{k+1}## and (using the fact that ##U_{k+1}## is a subspace) recognise that this gives a single element of ##U_{k+1}##. We also recognise that what is left is an element of ##span(U_1 \cup \dots \cup U_k)##.

Let me check my understanding :)

OK so let us start by taking ##v \in span(U_1 \cup \dots \cup U_{k+1})##. Based on the definition of span, ##v = \sum\limits_{j = 1}^{k+1} \beta_j u_j = \sum\limits_{j = 1}^{k} \beta_j u_j + \beta_{k+1} u_{k+1}##. The first and second terms equal (by definition) to ##span(U_1 \cup \dots \cup U_k)## and ##span(U_{k+1})## respectively. Since ##U_{k+1}## is a subspace, it is closed under linear combinations so ##span(U_{k+1}) \subseteq U_{k+1}##. But by definition of span, we also have ##U_{k+1} \subseteq span(U_{k+1})##. Hence ##U_{k+1} = span(U_{k+1})## and indeed we see that ## v \in span(U_1 \cup \dots \cup U_k) + U_{k+1}##
 
  • #11
PeroK
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Thanks for the clarification @PeroK !



Let me check my understanding :)

OK so let us start by taking ##v \in span(U_1 \cup \dots \cup U_{k+1})##. Based on the definition of span, ##v = \sum\limits_{j = 1}^{k+1} \beta_j u_j = \sum\limits_{j = 1}^{k} \beta_j u_j + \beta_{k+1} u_{k+1}##.
It's more general than than: $$v = \sum_{j = 1}^n \beta_jv_j$$where ##v_j \in U_1 \cup \dots \cup U_{k+1}##. Now, we let $$u_{k+1} = \sum_j \beta_jv_j$$ where we sum over all ##j## where ##v_j \in U_{k+1}## and let $$u = \sum_j \beta_jv_j$$ where we sum over all ##j## where ##v_j \notin U_{k+1}##. This gives us $$v = u + u_{k+1}$$, where ##u \in span(U_1 \cup \dots \cup U_k)## and ##u_{k+1} \in U_{k+1}##, hence:
$$v \in span(U_1 \cup \dots \cup U_k) + U_{k+1}$$
That can be used in your inductive step.

Note also that the base case for ##k = 1## is that we have ##span(U_1) = U_1##.
 

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