Prove that if a vector space $V$ has a basis of $n$ vectors, then every basis of $V$ must consist of exactly $n$ vectors?

Question

The proof in my textbook says:

Let $B_1$ be a basis of $ n $ vectors and $ B_2$ be any other basis of $V$.

(a)Because $ B_1$ is a basis and $B_2 $ is linearly independent, $B_2 $ has no more than n vectors.

(b)Because B$_2$ is a basis and $ B_1$ is linearly independent, $B_2$ has at least $n$ vectors. So $B_2$ has exactly $n $ vectors.

I understand (a) because I understand that if a vector space $V$ has a basis $ B = {b_1,...,b_n} $ then any set in $ V$ containing more than $n$ vectors must be linearly dependent.

But I don't understand the logic of (b).

SOS!

(b) is just the dual of (a). (a) states that B2 has no more vectors than B1 does. (b) states that B1 has no more vectors than B2 does. — Brian Cheung, Jun 08 '16 at 05:52
Another way would be to consider the position when $B_2$ has $m$ vectors. Then (a) is saying $m \le n$. By symmetry, (b) is saying $n \le m$ (just swap $B_1$ and $B_2$ in (a)). — Henry, Jun 08 '16 at 06:10
There is another proof you can try . Proving $B_1$ is isomorphic to $B_2$ will prove $dimB_1 = dimB_2$. — Shona, Jun 08 '16 at 07:12
It is the exact same logic both ways. If $B_2$ has fewer vectors than $B_1$ then the vectors in $B_1$ must have a dependency (and $B_1$ couldn't be a basis). — Doug M, Jun 09 '16 at 00:11
Does this answer your question? Any two bases of a finite dimensional vector space must have the same number of elements. — , Jul 27 '20 at 20:41

score 1 · Answer 1 · answered Jun 17 '19 at 14:24

Let $A$ be a basis with $m$ elements and $B$ be a basis with $n$ elements. Since both of them are basis of the vector space they have two properties i.e both of them span the vector space and both of them are linearly independent. Take $A$ and use the property that it spans the vector space and take $B$ with the property that it is linearly independent. Now try to see that the cardinality of a spanning set is always $\geq$ to the cardinality of a spanning set. So from here you get $m\geq n$. Now reverse the roles of $A$ and $B$ to get $m\leq n$.

score 0 · Answer 2 · answered Jun 08 '16 at 09:34

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The logic goes somehow like this: suppose $B_1$ has less elements than $B_2$. Lets say $B_1=\{b_1,\ldots,b_m\}$ and $B_2=\{v_1,\ldots,v_n\}$ and $m<n$. Now you can replace $m$ elements in $B_2$ by $B_1$ and still have a set of linearly independent set of vectors. Say $B'=\{b_1,\ldots,b_m,v_{m+1},\ldots,v_n\}$.

But now since $B_1$ was a basis you have that $\sum a_i b_i=v_{m+1}$ for some $a_i$ in your field. But this means

$$ \sum_{i=1}^m a_i b_i -v_{m+1}=0, $$

contradicting that $B'$ is linearly independent.

answered Jun 08 '16 at 09:34

Maik Pickl

2,345

I think your proof needs to be reworded since you're showing that your claim "Now you can replace $m$ ... and still have a LI set of vectors." is false. Properly written this is the correct argument. – user347489 Jul 30 '19 at 10:00
@user347489 The statement is certainly not false. Its called the Steinitz exchange lemma: https://en.m.wikipedia.org/wiki/Steinitz_exchange_lemma – Maik Pickl Aug 18 '19 at 21:21
You're right, however it's important to mention that the claim holds only up to reordering the elements of $B_2$. The way it's written this is not clear. – user347489 Aug 18 '19 at 21:34

score 0 · Answer 3 · edited Jul 30 '19 at 09:29

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$a.~~$ $B_2~$ has no more than cardinality $~B_1~$ vectors.

$b.~~$ $B_1~$ has no more than cardinality $~B_2~$ vectors.

Same logic when you annihilate $~n~$.

edited Jul 30 '19 at 09:29

nmasanta

9,640

answered Jun 09 '16 at 00:00

Jacob Wakem

2,420

score 0 · Answer 4 · answered Jul 30 '19 at 10:20

We can label the two bases whichever way round we like, so let's label them so that $|B_1| \le |B_2|$.

$B_1$ is a basis, so if $|B_2|$ were strictly greater than $|B_1|$ then $B_2$ would not be linearly independent. But $B_2$ is also a basis, so it must be linearly independent.

$\Rightarrow |B_1| = |B_2|$

score 0 · Answer 5 · answered Jan 02 '20 at 21:46

Let $B_{1}$ and $B_{2}$ two basis for the vector space $V$ suchs that $|B_{1}| = n$, $|B_{2}| = m$ with $m < n$ then $V \simeq \mathbb{R}^{n}$ and $V \simeq \mathbb{R}^{m}$ so by transitivity $\mathbb{R}^{n} \simeq \mathbb{R}^{m}$ which implies that $n = m$ but this is a contradiction.

Prove that if a vector space $V$ has a basis of $n$ vectors, then every basis of $V$ must consist of exactly $n$ vectors?

5 Answers5