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Recently, I solved a problem that says-
If $V$ is a vector space over an infinite field. Prove that, V cannot be written as set-thoretic union of a finite number of proper subspaces.
But is this result true in case of finite field?
. I can't get such an example where a vector space over finite field can be written as union of finite number of proper subspaces.
Can anybody give such an example? Thanks for assistance in advance.

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  • $\begingroup$ See here for the theory on how many subspaces you need at minimum, $\endgroup$ – Jyrki Lahtonen Nov 4 '18 at 5:31
  • $\begingroup$ That thread is linked to many others with more discussion and variants. Search them! $\endgroup$ – Jyrki Lahtonen Nov 4 '18 at 5:41
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The answer is yes; if $V$ is a finite vector space (so over a finite field and of finite dimension), then it has only finitely many elements. Let $\langle v\rangle$ denote the span of a vector $v\in V$. Then clearly $$V=\bigcup_{v\in V}\langle v\rangle,$$ because $v\in\langle v\rangle$ for every $v\in V$, and the union is finite because $V$ is finite. Hence $V$ is the union of finitely many proper subspaces (if $\dim V>1$, otherwise the subspaces aren't proper).

For a very concrete example, consider $\Bbb{F}_2^2$, a $2$-dimensional vector space over the finite field $\Bbb{F}_2$ of two elements. Then \begin{eqnarray*} \Bbb{F}_2^2&=&\{(0,0),(1,0),(0,1),(1,1)\}\\ \bigcup_{v\in\Bbb{F}_2^2}\langle v\rangle&=&\{(0,0)\}\cup\{(0,0),(1,0)\}\cup\{(0,0),(0,1)\}\cup\{(0,0),(1,1)\}. \end{eqnarray*}

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    $\begingroup$ But is the result true when the vector space is of infinite dimension? The vector space may not be finite. Example: set of all polynomials over $\Bbb{Z}_2$ $\endgroup$ – Biswarup Saha Nov 3 '18 at 19:17
  • $\begingroup$ Servaes, if $V$ is not finite dimensional then your method will not work, because in that case $V$ has infinitely many elements. $\endgroup$ – Biswarup Saha Nov 5 '18 at 6:59
  • $\begingroup$ What if the vector space is one-dimensional? Then the span of v is not a proper subspace. $\endgroup$ – odnerpmocon Nov 27 '19 at 11:34
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It seems to me that the case of an infinite dimensional space is creating extra headache. Spelling out an example to show what can be done.

Consider the field $\Bbb{F}_2$. Let $V=\Bbb{F}_2[x]$ be the space of univariate polynomials. Let $U$ be the subspace of polynomials with zero constant and linear terms. Consider the following three subspaces of $V$: $$ V_1=\langle 1\rangle\oplus U=\{a_0+a_1x+a_2x^2+\cdots +a_nx^n\mid n\in\Bbb{N}, a_i\in\Bbb{F}_2, a_1=0\}, $$ $$ V_2=\langle x\rangle\oplus U=\{a_0+a_1x+a_2x^2+\cdots +a_nx^n\mid n\in\Bbb{N}, a_i\in\Bbb{F}_2, a_0=0\}, $$ and $$ V_3=\langle 1+x\rangle\oplus U=\{a_0+a_1x+a_2x^2+\cdots +a_nx^n\mid n\in\Bbb{N}, a_i\in\Bbb{F}_2, a_0=a_1\}. $$ It is clear that every polynomial of $V$ belongs to at least one of the subspaces $V_1,V_2,V_3$ according to what its constant and linear terms look like. Therefore $$V=V_1\cup V_2\cup V_3.$$


What just happened?

Consider the quotient space $V/U$. It is 2-dimensional. Its non-zero elements are the cosets $1+U,x+U$ and $1+x+U$. So we can write $V/U$ as a union of three 1-dimensional subspaces following the recipe of Servaes' answer each spanned by one of those cosets. The resulting subspaces are exactly the spaces $V_i/U, i=1,2,3$. No wonder that $V_1,V_2,V_3$ cover all of $V$!

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