I am having a difficulty setting up the proof of the fact that two bases of a vector space have the same cardinality for the infinite-dimensional case. In particular, let $V$ be a vector space over a field $K$ and let $\left\{v_i\right\}_{i \in I}$ be a basis where $I$ is infinite countable. Let $\left\{u_j\right\}_{j \in J}$ be another basis. Then $J$ must be infinite countable as well. Any ideas on how to approach the proof?
Linear Algebra – Proof That Two Bases of a Vector Space Have the Same Cardinality
linear algebravector-spaces
Related Solutions
Yes it is true independent of the cardinality of the bases of the vector spaces. Use the universal property: Let $T$ be any vector space over the given field.
A bilinear map $V \times W \to T$ is uniquely determined by the images of the pairs $(v_i,w_j)_{i \in I,j \in J}$, so we get
$$Bil(V \times W, T) = Abb(\{(v_i,w_j), i \in I, j \in J\},T)= Abb(\{v_i \otimes w_j, i \in I, j \in J\},T)$$
By the universal property $V \otimes W$ is the vector-space with the property
$$Hom(V \otimes W,T) = Bil(V \times W, T)$$
, so we obtain
$$Hom(V \otimes W,T) = Abb(\{v_i \otimes w_j, i \in I, j \in J\},T)$$
, which precisely states that $\{v_i \otimes w_j, i \in I, j \in J\}$ is a basis of $V \otimes W$.
Let us precise this:
Given a map in $f \in Abb(\{v_i \otimes w_j, i \in I, j \in J\},T)$, which means that we are given $f(v_i \otimes w_j)$ for all $i \in I, j \in J$, we have to show that this extends uniquely to a map $F:V \otimes W \to T$.
We define the bilinear map $\hat f: V \times W \to T$ by $\hat f(v_i,w_j) := f(v_i \otimes w_j)$. By the universal property, this gives us a unique map $F:V \otimes W \to T, F(v \otimes w)=\hat f(v_i,w_j)=f(v_i \otimes w_j)$. Hence this is the desired unique map.
The uniqueness of $F$ is trivial anyway, because $\{v_i \otimes w_j, i \in I, j \in J\}$ is clearly a set of generators. We only need to show the existence of $F$ here (And this corresponds to the linear independence).
If $V$ is a vector space over the field $F$, and $B$ is a basis for $V$ then every element in $v$ is the unique combination of finitely many elements from $B$ with coefficients from $F$.
So each $v\in V$ can be represented as a finite subset of $B\times F$ which is a function. How? $v(b)=\alpha$ if $b$ is a basis element which has a nonzero coefficient, $\alpha$, in the linear combination of $v$ from $B$.
How many such functions are there? Now let's put into play the assumption that $V$ is infinite.
At least $B\times (F\setminus\{0\})$, since each singleton $\{\langle b,\alpha\rangle\}$ corresponds to the vector $\alpha\cdot b$. But if $X$ is infinite, the set of all finite subsets of $X$ has the same cardinality as $X$. So we get a bijection using Cantor-Bernstein. And since we're talking about infinite sets, we can consider $F$ and not $F\setminus\{0\}$.
So we get that $|V|=|B\times F|=|B|\cdot|F|$. But multiplying two non-zero cardinals, at least one of them is infinite, we get that $|B|\cdot|F|=\max\{|B|,|F|\}$.
In particular, if $|V|>|F|$ then $|V|=|B|$, or in other words $\dim V=|V|$.
So now if we want to apply this to $\Bbb R$ as a vector space over $\Bbb Q$, what do we get?
Best Answer
In spirit, the proof is very similar to the proof that two finite bases must have the same cardinality: express each vector in one basis in terms of the vectors in the other basis, and leverage that to show the cardinalities must be equal, by using the fact that the "other" basis must span and be lineraly independent.
Suppose that $\{v_i\}_{i\in I}$ and $\{u_j\}_{j\in J}$ are two infinite bases for $V$.
For each $i\in I$, $v_i$ is in the linear span of $\{u_j\}_{j\in J}$. Therefore, there exists a finite subset $J_i\subseteq J$ such that $v_i$ is a linear combination of the vectors $\{u_j\}_{j\in J_i}$ (since a linear combination involves only finitely many vectors with nonzero coefficient).
Therefore, $V=\mathrm{span}(\{v_i\}_{i\in I}) \subseteq \mathrm{span}\{u_j\}_{j\in \cup J_i}$. Since no proper subset of $\{u_j\}_{j\in J}$ can span $V$, it follows that $J = \mathop{\cup}\limits_{i\in I}J_i$.
Now use this to show that $|J|\leq |I|$, and a symmetric argument to show that $|I|\leq |J|$.
Note. The argument I have in mind in the last line involves some (simple) cardinal arithmetic, but it is enough that at least some form of the Axiom of Choice may be needed in its full generality.