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Thanks for looking at my question!


I: $\{b_1,...,b_n\}$ linear independent vectors and $\{c_1,...,c_n\}$ linear independent vectors.

II: $\{b_1,...,b_n\}$ span subspace spanned by $\{c_1,...,c_n\}$

Does $\{c_1,...,c_n\}$ necessarily span the subspace spanned by $\{b_1,...,b_n\}$? Note that dimension of B = dimension of C = n.

I'm pretty sure it does, but am not sure how to prove it. Also, is there any better way to write this? It's certainly less than poetic.

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If $n = \aleph_0$, this fails. Consider $k\mathbb{Z}$ - a vector space over the field $k$ spanned by integer numbers. Its subspace $k(2\mathbb{Z})$ has the same dimension, yet it's a proper subspace. – Alexei Averchenko Dec 12 '12 at 8:35
@AlexeiAverchenko In mathematics, $n$ tends to stand for natural numbers. But you might like the sequence $(n_\epsilon)$ that goes to zero as $\epsilon$ goes to infinity. – Michael Greinecker Dec 12 '12 at 9:07
@Alexei: Not only does $n$ usually stand for natural numbers, the notation $\{b_1,\dotsc,b_n\}$ almost always stands for a finite set, and if you wanted it to stand for something else then $n$ would have to be an appropriate ordinal. – joriki Dec 12 '12 at 9:14
The question seems to be tautologiacally true. If I read point II, it says the subspace spanned by $\{b_1,\ldots,b_n\}$ (call it $V_1$) equals the subspace spanned by $\{c_1,\ldots,c_n\}$ (call it $V_2$). And you are asking whether $V_2=V_1$? Well if $V_1=V_2$, then obviously "yes". Am I missing something? – Marc van Leeuwen Dec 12 '12 at 10:26
To add to my previous comment, with the above definitions you statement II could conceivably be interpreted as affirming any one of the statements $V_1\subseteq V_2$ (the $b_i$ span a subspace of the subspace $V_2$ spanned by the $c_i$), $V_1\supseteq V_2$ (the $b_i$ span at least $V_2$), or $V_1=V2$ (the $b_i$ span exactly the subspace $V_2$ spanned by the $c_i$). I think the last reading is what it says, but maybe not what was intended (given the number of answers people have read your mind better than I can; I feel stupid). Please clarify. – Marc van Leeuwen Dec 12 '12 at 10:49
up vote 8 down vote accepted

Yes, it does. If it didn't, there'd be a vector $b$ spanned by $\{b_1,\dotsc,b_n\}$ independent of $\{c_1,\dotsc,c_n\}$, so $\{c_1,\dotsc,c_n,b\}$ would span an $(n+1)$-dimensional space, contradicting the fact that $\{b_1,\dotsc,b_n\}$ spans that space.

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Only if the dimension is finite. – Alexei Averchenko Dec 12 '12 at 8:32

On a more poetic way to express this. I'll take a shot :) Below I'll consider both finite-dimensional and infinite-dimensional vector spaces over an arbitrary field $k$. You can think $k = \mathbb{R}$ if you like, it's not crucial.

Consider a vector space $V$ over the field $k$. There is a function $S \mapsto kS$ that takes each subset $S \subset V$ to its span $kS = \operatorname{span}S$ - the minimal subspace of $V$ containing $S$, that is the set of all elements that can be expressed as finite linear combinations of elements of $S$.

Interesting facts about this function:

  1. $S \subset kS$ for any $S$,
  2. if $T \subset S$, then $kT \subset kS$,
  3. $k(kS) = kS$ for any $S$.

These properties mean that this function is a closure operator. I admit I do not know much about closure operators except for topological closure operators, so it's just an aside in my answer :)

What's interesting about this function, though, is that it allows us to restate things about linear dependence in a more concise way.

Definition: a subset $S \subset V$ is called linearly dependent if there exists a proper subset $T \subsetneq S$ such that $kT = kS$. Otherwise we call it linearly independent.

Proposition: if a subset $S \subset V$ is linearly independent, then all $T \subsetneq S$ are also linearly independent.

Proof: Suppose there is $Q \subsetneq T$ such that $kQ = kT$. Then $$k(Q \cup S \setminus T) = k(kQ \cup k(S \setminus T)) = k(kT \cup k(S \setminus T)) = k(kS) = kS,$$ which contradicts the premise.

Consider all linearly independent subsets of $V$ ordered by inclusion.

Proposition: There is at least one maximal subset: one that's not included in any other.

Proof: Consider a chain $S_1 \subset S_2 \subset \ldots$, where each $S_i$ is linearly independent. Then $S = \bigcup_i S_i$ is also linearly independent: suppose $T \subsetneq S$ has the property $kT = kS$, then $T \cap S_i \subsetneq S_i$ has the property $kS_i = kS \cap kS_i = kT \cap kS_i = k(T \cap kS_i)$, thus by linear independence of $S_i$ we have $T \cap S_i = S_i$ for all $i$. But then $T = \bigcup_i (T \cap S_i) = \bigcup_i S_i = S$, which contradicts the premise. Therefore $S$ is indeed linearly independent. Apply the Zorn lemma to complete the proof.

Proposition: if $S \subset V$ is a basis of $V$, then $kS = V$.

Proof: Suppose $v \in V \setminus kS$. Then $S \subsetneq S \cup \{v\}$, and $S \cup \{v\}$ is linearly independent, which contradicts the maximality of the basis.

Definition: we call such a maximal linearly independent subset $S \subset V$ a basis of $V$.

Proposition: Let $S$ be a basis of $V$, and $U$ be a vector space over $k$. Then for every function $f: S \to U$ there is a unique linear operator $\varphi: V \to U$ such that for each $s \in S$ we have $\varphi(s) = f(s)$.

Proof: For any (finite) linear combination $v = a_1 s_1 + \ldots + a_k s_k$ we have $\varphi(v) = a_1 f(s_1) + \ldots a_k f(s_k)$, thus $f$ is uniquely defined on $kS = V$.

UPD: looks like the dimension theorem cannot be easily recast into these terms. Oh well :(

It gets even more poetic once we bring in category theory :)

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The following are for finite dimensional vector spaces.

If you don't know, prove:

  • $\dim(\text{span}(S))\leq |S|$ with equality iff $S$ is linearly independent, where $|S|$ is the cardinality of $S$.
  • If $U\leq V$ then $U=V \iff \dim(U)=\dim(V)$.

Denote $B=\{b_1,b_2,\ldots,b_n\}$ and $C=\{c_1,c_2,\ldots,c_n\}.$
Since $\text{span}(B)\leq\text{span}(C)$ and $\dim(\text{span}(B))=\dim(\text{span}(C)) \Longrightarrow \text{span}(B)=\text{span}(C).$

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