It's enough for say that does not exist $Span(S∪T)=Span(S)⨁Span(T)$?

Question

Given $S,T ⊆ V$ so that: $S,T\neq ∅$.

It's will be true to say that:

If $S∩T=∅$ so $Span(S∪T)=Span(S)⨁Span(T)$?

I think that not, because the elements belongs to $Span(S∪T)$ it's not like the elements that belongs to $Span(S)⨁Span(T)$. But I don't sure about it.

The elements are fine, look closely at the $\oplus$. – eranreches Dec 17 '17 at 14:22 — eranreches, Dec 17 '17 at 14:22

Mariah · Accepted Answer · 2017-12-17T14:34:29.127

3

No, take $S = \{(1,2)\}$, $T = \{(2,4)\}$. Then $S \cap T = \emptyset$, but $span(S \cup T) = span(S), span(T)$

In general, you can't even take the direct sum of $S$ and $T$ here because their corresponding spanning subspaces are equal.

The reason you can't is because a representation wouldn't be unique. For instance $(1,2) = (1,2) + 0(2,4) = 0(1,2) + \frac{1}{2}(2,4)$.

See the second answer here:

Difference between sum and direct sum

edited Dec 17 '17 at 14:34

answered Dec 17 '17 at 14:24

Mariah

3,880

Can you explain why $Span(S∪T) \neq Span(S)⨁Span(T)$ in your example? – Master Question Dec 17 '17 at 14:30
@MasterQuestion is it more clear now? – Mariah Dec 17 '17 at 14:35
Yes, by the way, you have example that $S$ and $T$ are without same base so that it's will not be true? – Master Question Dec 17 '17 at 15:26

score 0 · Answer 2 · answered Dec 17 '17 at 14:24

0

It's not true. For example, take $T=\{(1,0)\}$ and $S=\{(2,0)\}$. They are disjoint, but the basis of their spans are equal, therefore the intersection of their spans is not equal to $\{0\}$, so it does not agree with the definition of $\bigoplus$.

answered Dec 17 '17 at 14:24

The Phenotype

5,300

It's enough for say that does not exist $Span(S∪T)=Span(S)⨁Span(T)$?

2 Answers2