An analogous equivalence to one of linear independence

Question

In a finite dimensional vector space $V$, a subset $S \subseteq V$ is said to be linearly dependent if there is a nontrivial solution $(a_i)$ to the equation

$$ \sum_{\mathbf{v_i}\in S}a_i\mathbf{v_i} = \mathbf{0} $$

An equivalent characterisation is that a set $S\subseteq V$ is linearly dependent if there is some $\mathbf{v}\in S$ such that

$$ Span(S) = Span(S/\{\mathbf{v}\}) $$ This allows one to prove that if the cardinality of the set is greater than the dimension of the space spanned by the set, then the set must be linearly dependent:

$$ |S| > dim(span(S)) $$

But the span of $S$ is nothing but the direct sum of the subspaces spanned by the elements of $S$.

This motivates the question, is there an analogous concept to linear independence but for subspaces instead of simply vectors?

Define by the $span$ of a set of subspaces the direct sum of all those subspaces.

Specifically I want to know if we have a set of subspaces $\mathcal{S}$ such that

$|\mathcal{S}|>dim(span(\mathcal{S}))$

Do we also necessarily have that there is some $T\in \mathcal{S}$ such that

$ span(\mathcal{S}) = span(\mathcal{S}/\{T\}) $

The context of this question is that if this second equivalence were true, it would allow for a fairly elegant proof of an Olympiad problem, co concerning sequences of square matrices that have pairwise $0$ products, but where the square of any matrix is non $0$. Initially one might think that this problem is trivially true since one can associate a subspace with a matrix, but linear combinations of these matrices are not the same as direct sums of the corresponding subspaces.

Answer 1

First, I would advise using the existing term "direct sum" rather than overloading a different term.

I believe the answer to your question is yes. Hint: prove the contrapositive by induction on $\#S$.

Answer 2

Inspired by the hint provided by @Greg Martin we prove the contrapositive by induction on $\#\mathcal{S}$.
The contrapositive of the above statement is

$$ \forall_{T\in\mathcal{S}},\bigoplus \mathcal{S} \neq \bigoplus (\mathcal{S}/\{T\}) \\ \implies \#\mathcal{S} \leq dim(\bigoplus \mathcal{S}) $$

For the base case, if $\#\mathcal{S} = 1$ then

$\#\mathcal{S} = dim(\bigoplus\mathcal{S})$ *

So the base case is proven.

We assume that for some $k\in\mathbb{N}$, and for any set $\mathcal{S}$ such that $\#\mathcal{S}=k$ the following is true:

$$\forall_{T\in S}, \bigoplus\mathcal{S} \neq \bigoplus(\mathcal{S}/\{T\}) \implies \\ \#\mathcal{S} \leq dim(\bigoplus\mathcal{S}) $$

We then want to prove that for any $\mathcal{S}$ such that $\#\mathcal{S} = k+1 $, the above implication is true. So let $\mathcal{S}$ be a set with $\#\mathcal{S}=k+1$ and assume

$ \forall_{T\in S}, \bigoplus\mathcal{S} \neq \bigoplus(\mathcal{S}/\{T\}) $

Then it is clearly any subset of $\mathcal{S}$ with $k$ elements satisfies the conditions of the inductive hypothesis, since if we assume that there is some subset $\mathcal{S'}$ with $k$ elements for which

$ \exists_{T\in \mathcal{S'}},\bigoplus{S'}=\bigoplus(\mathcal{S'}/\{T\}) $

Then we can choose that same $T$ and find that

$ \exists_{T\in \mathcal{S}},\bigoplus{S}=\bigoplus(\mathcal{S}/\{T\}) $

Which contradicts the assumption of our. Therefore we can apply the conclusion of the inductive hypothesis to conclude that for any subset $\mathcal{S'}$ of $\mathcal{S}$ with $k$ elements,

$\dim(\bigoplus\mathcal{S'})\geq k $

It is then rather straightforward to show that $\dim(\bigoplus\mathcal{S})>k$ since if this were not the case, we have a subset $\mathcal{S'}$ of size $k$ for which

$\bigoplus\mathcal{S'}=\bigoplus\mathcal{S}$

Which contradicts our assumption.

Therefore

$\dim(\bigoplus\mathcal{S})\geq k+1 $

And the proposition follows by induction.

*It was not mentioned in the question, but one may assume that the elements of $\mathcal{S} $ are non-trivial subspaces, the reason for this is clear in the context of the olympiad question, but the exception of the trivial subspace can easily be dealt with as it is done with vectors.

Answer 3

This post is aimed at showing how this proposition completes the proof of the olympiad problem.

Problem: Let a sequence $(A_i)_{i\leq k}$ of $n\times n$ real matrices be called "good" if $A_i^2\neq 0$ and $i\neq j \implies A_iA_j = 0$. Prove that a good sequence has at most $n$ elements.

Proof: Denote by $R(A_i)$ and $C(A_i)$ the row- and column spaces of the matrix $A_i$. Note that if $A_iA_j = 0$ then $R(A_i)$ is orthogonal to $C(A_j)$, which we will denote by $R(A_i)\perp C(A_j)$ Where orthogonal here means one subspace is a subset of the orthogonal complement of the other.

Now assume we have a good sequence, $(A_i)$ of length $n$. We first prove a lemma which is that

$ \dim(\bigoplus_{i\leq n}R(A_i)) = n$.

For the sake of contradiction, assume that for some $0<k\leq n$,

$ \dim(\bigoplus_{i\leq n}R(A_i)) = n-k$

Then By the original proposition of this post, since we have

$\#(A_n) > \dim(\bigoplus_{i\leq n}R(A_i)) $

We can conclude that there is some $A_j$ such that

$\bigoplus_{i\leq n}R(A_i) = \bigoplus_{i \leq n, i\neq j} R(A_i) \hspace{0.5cm} (1) $

In other words there is some matrix $A_j$ that can be removed from the sequence without changing the direct sum of the subspaces. But since the sequence $(A_n)$ is good, we know that $A_iA_j = 0$ for $i\neq j$, which means that for every $i \neq j$, $R(A_i)\perp C(A_j)$.

This in turn implies

$C(A_j)\perp \bigoplus_{i\leq n, i\neq j} R(A_i) $

But

$R(A_j) \subseteq \bigoplus_{i\leq n, i\neq j} R(A_i) \mbox{ by (1)}$

So

$C(A_j)\perp R(A_j) $

But this means that $A_j^2 = 0$, contradicting the assumption that the sequence $A_n$ was good. So we must have that $k=0$ or

$ \dim(\bigoplus_{i\leq n}R(A_i)) = n$.

This completes the proof of the lemma. To prove the question statement, assume we are given a good sequence, $(A_i)$ of length $k>n$. Then by the preceding lemma,

$ \dim(\bigoplus_{i\leq n}R(A_i)) = n$

(The row spaces of the first $n$ matrices span the entire space $\mathbb{R}^n$)

But since $C(A_{n+1}) \perp R(A_{i}) $ for $i\leq n$

We get that

$C(A_{n+1}) \perp\bigoplus_{i\leq n}R(A_i) = \mathbb{R}^n$

However this can only be if

$C(A_{n+1}) = \{\mathbb{0}\} $

But then $A_{n+1}^2=0$, contradicting the assumption that the sequence was good, thus a good sequence cannot have more than $n$ elements.