# Show that $\{v_1,v_2,\dots,v_n\}$ is a basis of a vector space iff a chain of subspaces is complete.

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.

• Does $V$ have dimension $n$? Jun 30, 2020 at 16:10
• @paulinho I would assume so, since the basis has $n$ elements. Jun 30, 2020 at 16:14
• 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. Jun 30, 2020 at 17:09
• @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? Jun 30, 2020 at 17:35
• Yes, completeness is not required to prove independency.
– user598858
Jun 30, 2020 at 18:11

## 1 Answer

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 $$dim(V_{i}/V_{i-1}) = dim(V_{i}/W) + dim(W/V_{i-1})\geq 2$$

• 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 $dim(V/V')=1$. Jun 30, 2020 at 20:30