Subspaces and Direct Sums

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[carmine pulsar]

you have to be more specific

a) is false under these assumptions

take U_i = V for every i, for example

you have forgotten something somewhere in here

@pseudo mantle

[pseudo mantle]

It is U_1, . . . , U_n ⊊ V

[carmine pulsar]

doesn't render here at least

either way, what are you stuck with then?

[pseudo mantle]

I don't know how to start the proof.

[carmine pulsar]

can you justify this part?

what's the opposite statement and what would happen if that was the case?

why would there be a contradiction?'

contradiction with what?

let's assume n=2, so you have two subspaces (of equal dimension)

call them U,W

if you assume this

then U subset W and W subset U

so U=W

can you extend this idea to n subspaces?

[pseudo mantle]

Let’s consider the case n=2 with two subspaces U and W. If we assume that U ⊆ W and W ⊆ U, it follows that U = W because each subspace is contained in the other. Now, if we generalize this idea to n subspaces U1, U2, ..., Un, and assume for every i that Ui ⊆ ⋃(j ≠ i) Uj, this would imply that all subspaces are contained in each other. Eventually, this would collapse to U1 = U2 = ... = Un, meaning all subspaces are identical. However, if this were true, their union would just be one subspace with the same dimension as Ui, contradicting the fact that V has a larger dimension. So, there must exist some Ui that is not fully contained in ⋃(j ≠ i) Uj.

[carmine pulsar]

 Now, if we generalize this idea to n subspaces U1, U2, ..., Un, and assume for every i that Ui ⊆ ⋃(j ≠ i) Uj, this would imply that all subspaces are contained in each other.

why?

can you demonstrate this with three subspaces?

U1,U2,U3, for example

[pseudo mantle]

If we assume that for each ( U_i \subseteq \bigcup_{j \neq i} U_j ), meaning that each subspace ( U_i ) is contained in the union of the other subspaces, this leads to increasing overlap between the subspaces. For example, ( U_1 \subseteq U_2 \cup U_3 ) means every element of ( U_1 ) is either in ( U_2 ) or in ( U_3 ). Repeating this for all ( U_i ) leads to such overlap that eventually all subspaces must be identical, which contradicts the assumption of a higher-dimensional space with distinct subspaces. In particular, subspaces of equal dimension that are contained in the union of the others would collapse into one, meaning the union cannot cover the entire space ( V ). This creates a contradiction when ( V ) has a higher dimension, showing that there must exist at least one subspace ( U_i ) that is not fully contained in the union of the others.

[glossy field]

Céline

[carmine pulsar]

you are rephrasing the problem, why does this "overlap" happen the way you say it does?

[pseudo mantle]

When we assume that ( U_i \subseteq \bigcup_{j \neq i} U_j ), we are saying that every element of ( U_i ) must lie within the union of the other subspaces. This means that any vector in ( U_1 ), for example, must be in ( U_2 ) or ( U_3 ) (or both). Similarly, every vector in ( U_2 ) must lie in ( U_1 ) or ( U_3 ), and the same holds for ( U_3 ).
Since each subspace is contained in the union of the other subspaces, this creates a situation where the elements of one subspace must "share" vectors with the others. For example, any vector in ( U_1 ) must also be in ( U_2 ) or ( U_3 ), and any vector in ( U_2 ) must also be in ( U_1 ) or ( U_3 ). This means that the subspaces cannot be "disjoint" — they must intersect in some way.
If the dimensions of the subspaces are equal, and if each subspace is contained within the union of the others, the only way this can hold true is if the subspaces must coincide. To see why, consider that if ( U_1 \subseteq U_2 \cup U_3 ), then every vector in ( U_1 ) is either in ( U_2 ) or ( U_3 ). Similarly, vectors in ( U_2 ) and ( U_3 ) must lie within the other two subspaces. If this overlap continues, it forces ( U_1 ), ( U_2 ), and ( U_3 ) to end up being the same subspace. In the case of equal dimensions, this means that the union of the subspaces cannot cover a space larger than any individual subspace.
The above reasoning leads to a contradiction because if all subspaces were to collapse into one, then their union would no longer span the original vector space ( V ). This would violate the assumption that the union of these subspaces should cover ( V ), unless the subspaces are distinct and non-overlapping in a way that allows their union to span ( V ).

[glossy field]

Céline

[carmine pulsar]

ok better, you have included the fact they all have the same dimension

you make claims, but you don't give proofs

explain this part properly, for example

let's assume for now this is true

why is this true?

and why does it imply that this is true?

this one is easy assuming you know every linearly independent subset can be extended to a basis

[pseudo mantle]

To prove part (ii), we start with the assumption that ( x \in U_i \setminus \bigcup_{j \neq i} U_j ) and ( y \in V \setminus U_i ). This means that ( x ) is in ( U_i ), but not in the union of the other subspaces, and ( y ) is in ( V ) but not in ( U_i ). The set we need to consider is ( M_{x, y} = { y + \lambda x \mid \lambda \in K^\times } ), where ( K^\times ) represents the nonzero elements of the field ( K ).

Since ( x \neq 0 ), and for each nonzero scalar ( \lambda \in K^\times ), the vector ( y + \lambda x ) is distinct from other vectors in the set, we see that ( M_{x, y} ) contains infinitely many distinct vectors. Specifically, for any two different values of ( \lambda_1 ) and ( \lambda_2 ) in ( K^\times ), the vectors ( y + \lambda_1 x ) and ( y + \lambda_2 x ) are distinct, because ( \lambda_1 \neq \lambda_2 ) implies ( \lambda_1 x \neq \lambda_2 x ), and thus ( y + \lambda_1 x \neq y + \lambda_2 x ).

Therefore, the set ( M_{x, y} ) has infinite cardinality, as it contains an infinite number of distinct vectors corresponding to all possible nonzero values of ( \lambda ).

[glossy field]

Céline

[carmine pulsar]

that is correct

if you want to be fancy, you can say that the map

$$ K^\times \to V,\quad \lambda\mapsto y+\lambda x$$

[glossy field]

aL

[carmine pulsar]

is injective

by assumption K is infinite field, so the set M(x,y) is infinite

[pseudo mantle]

To prove part (b), we need to show that for each ( i = 1, \dots, n ), there exists a subspace ( W \subseteq V ) such that ( U_i \oplus W = V ). This means that the subspace ( U_i ) has a complementary subspace in ( V ).

Since the subspaces ( U_1, U_2, \dots, U_n ) all have the same dimension, we can choose a basis for each ( U_i ). Let ( { x_1, x_2, \dots, x_k } ) be a basis for ( U_i ). Since all the ( U_i )'s have the same dimension, each of their bases will consist of the same number of vectors. We now extend this linearly independent set to a basis of ( V ), namely ( { x_1, x_2, \dots, x_k, x_{k+1}, \dots, x_m } ), where ( m = \dim(V) ).

Next, we define the subspace ( W ) as the span of the vectors ( { x_{k+1}, \dots, x_m } ). These vectors are linearly independent from the vectors in ( U_i ), so ( W ) is complementary to ( U_i ). This means that every vector in ( V ) can be uniquely written as the sum of a vector from ( U_i ) and a vector from ( W ).

Thus, we have constructed a subspace ( W \subseteq V ) such that ( U_i \oplus W = V ) for each ( i ).

[glossy field]

Céline

[carmine pulsar]

nono, read b) again

there exists W such that for all i W is complementary to U_i

for example, in R^2 you can take the x-axis and y-axis as your subspaces of equal dimension

the line y=x is a subspace that is complementary to the x-axis and y-axis

you can do this with R^3 and axes or planes too

but now prove this works in general

[glossy field]

Céline

[carmine pulsar]

Now, for each subspace ( U_i ), we can extend its basis in a similar way, adding vectors that are linearly independent from the vectors in ( U_i ). The set of these additional vectors will form the same subspace ( W ).

Not true

E.g in R^2 we have x-axis and y-axis.
both {(1,0), (1,1)} and {(0,1), (-1,1)} are bases of R^2 but span {(1,1)} is not equal to span {(1,-1)}

you have to pick a basis for W in a certain way

and you can't have it depend on the index i

you have to involve all the subspaces U_i at once

remember this part

you use this fact in b)

[limber helm]

@carmine pulsar I'm a bit confused, there's nothing preventing us from having U1 = U2 = ... = Un. In that case. Statement 1.a)i) would be false ...

[limber helm]

This doesn't work if all U_i are identical, and there's nothing stating it can't be the case in the problem

[pseudo mantle]

It is true because of this. Something went wrong with the translation, that's why the equals sign is not crossed out

[limber helm]

Yeah but besides that,
Assume V = Vect (e1,e2,e3). You can take
U1 = U2 = Vect (e1,e2).
Then :
U1, U2 ⊆ V
but : U1 = U2

In order for this to work. You need :
Ui ≠ Uj for each (i,j) in [1,n]

Am I missing something ?

[glossy field]

Céline

[limber helm]

Yes, but that won't work if all the Ui you choose are identical. And there's nothing prohibiting that in the problem statement.

[carmine pulsar]

well you can assume they are not identical

A,B,B,C is the same list of subspaces as A,B,C

and if there is only one distinct subspace in that list then there's nothing to prove

[limber helm]

Yes but that's not the problem statement. They didn't prohibit it, and this makes this statement false.

[carmine pulsar]

that part is not the problem

[limber helm]

Nothing prohibits me from choosing them to be identical. And in that case, this becomes false

@pseudo mantle, if the problem is in french. Could you share the original problem statement ?

[carmine pulsar]

fair enough, assume there are at least two distinct subspaces

it's a minor issue

[limber helm]

Oh alright 🙂

[carmine pulsar]

the important assumption is that they have the same dimension

[limber helm]

@pseudo mantle, can you send the original problem statement (in German)

Thanks :), are you still trying to solve it ?

[pseudo mantle]

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