(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

This is the exact phrasing form Linear Algebra Done Right by Axler:

Prove that the union of three subspaces of V is a subspace of V if and only if one of the subspaces contains the other two. [This exercise is surprisingly harder than the previous exercise, possibly because this exercise is not true if we replaceFwith a field containing only two elements.]

3. The attempt at a solution

Let U_{a},U_{b},U_{c}be subspaces of the vector space V. Let a∈U_{a}, b∈U_{b}, c∈U_{c}and Let U= U_{a}∪ U_{b}∪U_{c}. For U to be a subspace:

1.The identity vector must exist: it's there since the U is a union of subspaces.

2.Closed under addition and scalar multiplication:

a,b,c∈U→a+b,b+c,a+c,a+b+c∈U

Now consider the following cases:

case 1:

Let it be that for all the subsets

x+y∈U_{x}or x+y∈U_{y}where, x,y∈{a,b,c} and x≠y. (I'll refer to this as "condition")

Let's take as an example: a+b∈U_{a}, b+c∈U_{b}, a+c∈U_{c}

a+b∈U_{a}→ b∈U_{a}

b+c∈U_{b}→ c∈U_{b}

a+c∈U_{c}→ a∈U_{c}

Now a+b+c can belong to any U_{i}and that would imply that a,b,c belong to that U_{i}proving our proposition. (Or we can use the fact that the subsets are equivalent since they're contained in each other)

case 2:

Two of the subsets satisfy the previous condition but one does not.

Let's take as an example: a+b∈U_{a}, a+c∈U_{b}, b+c∈U_{c}

a+b∈U_{a}→ b∈U_{a}

b+c∈U_{c}→ b∈U_{c}

Now if a+b+c∈U_{a}or a+b+c∈U_{c}then our proposition is proved. However, if it belongs to U_{b}then nothing is implied. But by the implications above, b belongs to both U_{a}and U_{c}. Therefore:

U_{b}⊂ U_{c}and U_{b}⊂ U_{a}→ a+c∈U_{a}and a+c∈U_{c}. Proving our proposition.

case 3:

None of the subsets satisfy the previous condition so we have:

b+c∈U_{a}, a+c∈U_{b}, a+b∈U_{c}

consider ai+bj+ck where i,j,k∈F. ai+bj+ck∈U so it must belong to one of subsets. However, if i=j or j=k or i=k, then it is obvious that it would belong to one of the subsets;

for example: if i=j then (a+b)i+ck∈U_{c}by addition and scalar closure.

So to take the argument further, suppose that i≠j≠k. Then ai+bj+ck belongs to one of the subsets. For the sake of the argument let that subset be U_{a}. Then, since ai∈U_{a}, and (b+c)∈U_{a}we can do the following:

ai+bj+ck−(ai+(b+c)j)=c(k−j). Therefore c(k−j)∈U_{a}and this implies that c∈U_{a}.

Finally since a,c,b+c∈U_{a}→ b∈U_{a}completing the argument. This argument can be replicated for when ai+bj+ck∈U_{b}or U_{c}.

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My Main question is regarding whether this covered all the cases and that there was not any wrong assumptions in my proof. I've reviewed multiple times but I still have my doubts. Also, I know that the statement is not precisely true, and that the cardinality of a field F must be greater than or eqU_{a}l the number of subsets for the statement to be true; however, this was not covered by the book yet so I didn't worry about it. I've written the proof very informally since it's "exhaustive" because it considers multiple cases and it would be difficult to write it more formally. Please let me know if a clarification is needed.

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# The union of three subspaces of V is a subspace of V

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