A fundamental fact about Linear Algebra

[caffeinemachine]

Hello MHB,
This is probably my first challenge problem which falls in the 'University Math' category.

$V$ is a vector space over an infinite field $F$, prove that $V$ cannot be written as a set theoretic union of a finite number of proper subspaces.

 

[caffeinemachine]

Nobody participated :(

Here's my solution:

Assume contradictory to the problem. Let $n$ be the minimum integer such that $V$ can be written as $V=V_1\cup\cdots\cup V_n$ where each $V_i$ is a proper subspace of $V$. Thus, \begin{equation*}\forall i,\exists x_i\in V \text{ such that } x_i\in V_j\iff j=i\tag{1}\end{equation*}Now consider $S=\{f_1x_1+\cdots+f_nx_n:f_i\in F\}$. Clearly this set is infinite, thus, by PHP, there is a $k$ such that $a,b\in V_k$ for distinct $a$ and $b$ in $S$. This contradicts $(1)$. Hence we achieve the required contradiction and the proof is complete.

 

[MarkFL]

Nobody participated :(

I would give everyone at least a week to participate, since not everyone checks in on a daily basis. Some may only have time once a week.

 

[caffeinemachine]

I would give everyone at least a week to participate, since not everyone checks in on a daily basis. Some may only have time once a week.

My bad then. Next time I'll wait for a week.

 

[CellCoree]

I would like to provide a response to this content by first acknowledging that linear algebra is a fundamental and essential branch of mathematics that has numerous applications in various fields such as physics, engineering, economics, and computer science. It deals with the study of vector spaces and linear transformations, and it has been extensively studied and applied for centuries.

In regards to the statement that $V$ cannot be written as a set theoretic union of a finite number of proper subspaces, I would like to provide a proof to support this fact. First, let us define what a proper subspace is. A proper subspace of a vector space $V$ is a subset $U$ of $V$ that is a vector space in its own right, but does not contain all the elements of $V$. In other words, $U$ is a proper subspace if it does not equal $V$.

Now, let us assume that $V$ can be written as a set theoretic union of a finite number of proper subspaces, say $U_1, U_2, ..., U_n$. This means that every element in $V$ can be written as a linear combination of elements in $U_1, U_2, ..., U_n$. However, this implies that $V$ is a finite-dimensional vector space, which contradicts the given fact that $F$ is an infinite field. This is because a finite-dimensional vector space over an infinite field $F$ has an infinite number of elements.

Furthermore, if $V$ is written as a finite union of proper subspaces, then there exists an element $v \in V$ that cannot be written as a linear combination of elements in $U_1, U_2, ..., U_n$. This element $v$ would then not belong to any of the proper subspaces, which contradicts the assumption that $V$ is the union of these subspaces.

Therefore, it can be concluded that $V$ cannot be written as a set theoretic union of a finite number of proper subspaces. This fundamental fact has important implications in linear algebra and serves as a reminder of the infinite nature of vector spaces over infinite fields.