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Let $V$ be a vector space over a field $F$. A chain $\{0\}=V_0\subseteq V_1\subseteq\dots\subseteq V_{n-1}\subseteq V_n=V$ of subspaces $V_1,V_2,\dots,V_{n-1}$ of $V$ is said to be complete if there is no subspace $W$ of $V$ such that $V_i\subsetneq W\subsetneq V_{i+1}$ for any $i=0,1,\dots n-1$.

Problem

Let $\{0\}=V_0\subseteq V_1\subseteq\dots\subseteq V_{n-1}\subseteq V_n=V$ be a chain of subspaces $V_1,V_2,\dots,V_{n-1}$ of a vector space $V$ over a field $F$. Let $v_1,v_2,\dots,v_n\in V$ such that $v_i\in V_i\setminus V_{i-1}$ for $i=1,2,\dots,n$. Show that $\{v_1,v_2,\dots,v_n\}$ forms a basis for $V$ if and only if the chain is complete.

I came across with this very new interesting problem (for me) mentioned above. Irrespective of completeness of the chain I could prove that $\{v_1,v_2,\dots,v_n\}$ is linearly independent. But to prove that it spans $V$, it requires the completeness of the chain, where I am stucked at. Please any one can help me with this problem. Thank you.

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  • $\begingroup$ Does $V$ have dimension $n$? $\endgroup$
    – paulinho
    Jun 30, 2020 at 16:10
  • $\begingroup$ @paulinho I would assume so, since the basis has $n$ elements. $\endgroup$ Jun 30, 2020 at 16:14
  • $\begingroup$ In that case it would have to be impossible to prove that the vectors are independent without the assumption of completeness, contrary to what OP claims they’ve done. $\endgroup$
    – paulinho
    Jun 30, 2020 at 17:09
  • $\begingroup$ @paulinho I don't see why completeness is needed--just $v_{i}\in V_{i}\setminus V_{i-1}$. Write $\lambda_{1}v_{1}+ \ldots +\lambda_{n}v_{n}=0$. If some $\lambda_{i}$ is nonzero then take $i$ maximal st $\lambda_{i}\neq 0$. Then $v_{i}$ is in the span of $\{v_{j} : j< i\}$ which is a subspace of $V_{i-1}$, contradiction. What am I missing? $\endgroup$
    – halrankard
    Jun 30, 2020 at 17:35
  • $\begingroup$ Yes, completeness is not required to prove independency. $\endgroup$
    – user598858
    Jun 30, 2020 at 18:11

1 Answer 1

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I am expanding on my original answer to help clarify the confusion in the discussion of the problem.

Assumptions. We have an arbitrary vector space $V$ (of unspecified dimension) over a field $F$ and a chain $\{0\}=V_{0}\subseteq V_{1}\subseteq \ldots \subseteq V_{n-1} \subseteq V_{n} = V$ of subspaces. We then pick vectors $v_{1}, \ldots, v_{n}$ such that $v_{i}\in V_{i}\setminus V_{i-1}$.

Note in particular that each $v_{i}$ is nonzero since $v_i$ is not in $V_0$.

Remark 1. $v_{1}, \ldots, v_{n}$ are linearly independent.

Proof. Suppose $\lambda_{1}v_{1}+ \ldots +\lambda_{n} v_n = 0$. If some $\lambda_{i}$ is nonzero then we may choose largest such $i$. But then $v_i$ is a linear combination of $v_{1}, \ldots , v_{i-1}$ contradicting our assumption that $v_{i}$ is not in $V_{i-1}$.

Main Claim. $\{v_{1}, \ldots, v_{n}\}$ is a basis iff the chain of subspaces is complete.

Proof.

$\Leftarrow$: Assume the chain is complete. By Remark 1 we only need to prove that $\{v_{1}, \ldots, v_{n}\}$ spans $V$. Note that for any $i$, $V_{i}$ has dimension $1$ over $V_{i-1}$. Since if not, then we can pick two vectors $v$ and $w$ in $V_{i}$ linearly independent over $V_{i-1}$. But then the space $W$ spanned by $V_{i-1}$ and $v$ is strictly between $V_{i}$ and $V_{i-1}$. So for any $i$, we have that $V_{i-1}$ is spanned by $V_{i-1}\cup\{v_{i}\}$ since $\{v_i\}$ must be a basis for $V_{i}$ over $V_{i-1}$ by the above conclusion. Then $V_{n}=V$ is spanned by $\{v_{1},\ldots v_{n}\}$ by induction. Another way to see this is to use the general formula $$ dim(V) = dim(V_{n}/V_{n-1}) + dim(V_{n-1}/V_{n-2}) + \ldots + dim(V_{1}/V_{0}) $$ We have shown that each summand on the right is $1$. So $dim(V) = n$ hence $\{v_{1}, \ldots , v_{n}\}$ is a basis.

$\Rightarrow$: Assume $\{v_{1}, \ldots, v_{n}\}$ is a basis. Then $dim(V)=n$ by Remark 1. Consider the same formula for $dim(V)$ in terms of the dimensions of the quotient spaces as above. The existence of the $v_{i}$'s ensures each $dim(V_{i}/V_{i-1})$ is at least $1$. So each of these is exactly $1$ since $dim(V)=n$. This forces the chain to be complete, since if $W$ is strictly between $V_{i-1}$ and $V_{i}$ then we would have $$ dim(V_{i}/V_{i-1}) = dim(V_{i}/W) + dim(W/V_{i-1})\geq 2 $$

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  • $\begingroup$ One of the main points of the whole situation is that if we have strictly nested subspaces $V'\subset V$ then there is no subspace $W$ strictly between $V'$ and $V$ iff $dim(V/V')=1$. $\endgroup$
    – halrankard
    Jun 30, 2020 at 20:30

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