# Proving a vector space is infinite-dimensional

Let $$S$$ be a vector space. Suppose that for any positive integer $$n$$, there exists a linearly independent subset $$S_n ⊆ V$$ of size $$n$$.

How do you show that $$S$$ is not finite dimensional but infinite-dimensional.

I understand that to be finite dimensional then it must have a finite then there exists a finite set that such at $$\operatorname{Span}(T)=S$$. So would it be right to prove by contradiction that there is not a finite spanning set that satisfies $$\operatorname{Span}(T)=S$$, and hence I could argue that the vector space $$S$$ is infinite dimensional.

How would I show that $$S$$ does not have a spanning set $$T$$, such that $$\operatorname{Span}(T)=S$$?

• If it was finite dimensional it would have a finite basis. Mar 5 '18 at 22:42

Try to prove the following statement:

If $V$ is a vectorspace and $A = \{x_1, \ldots, x_n\}$ is a subset of $V$ containing $n$ (different) vectors. Suppose $A$ is a generating set for $V$, then every subset of $V$ with more than $n$ elements is a linearly dependent subset.

Given: a vector space $V$ such that for every $n \in \{1, 2, 3, \ldots\}$ there is a subset $S_n$ of $n$ linearly independent vectors.

To prove: $V$ is infinite dimensional.

Proof: Let us prove this statement by contradiction: suppose that $V$ has finite dimension $k$. Fix a basis $\{v_1, \ldots, v_k\}$, then it is given that we can find a set $S_{k+1}$ of $k +1 < \infty$ vectors which are linearly independent. However, the above statement shows that the elements have to be linearly dependent, a contradiction. Hence it must be that $V$ is infinite dimensional.

• How can one assume that by assumption there is a set k+1<infinity elements such that the elements are linearly independent? Thanks Mar 6 '18 at 14:02
• @Mathamainia see edit. If this is not clear, I will try to explain :). Also, note that I denoted the vector space by $V$, since I think using $S$ would be confusing together with the $S_n$. Mar 6 '18 at 14:18
• Sorry would you mind please explaining Mar 6 '18 at 14:41
• @Mathamainia have you read what I have edited? The statement you try to prove says you have a vector space $V$ such that for each $n$ there is a subset $S_n$ of $n$ linearly independent vectors. You try to prove that $V$ is infinite dimensional. I prove this by contradiction, by assuming that $V$ has finite dimension and then use that it is given that I can find a subset with more vectors which are linearly independent. Mar 6 '18 at 14:43
• Surely any set Sk+1 of k+1<∞ vectors would be are linearly dependent by the theorem you stated above, and as we are given S is a linearly independent set of vectors then this causes contradiction and hence V is infinite dimensional. (rather than the other way) Mar 18 '18 at 15:42

I will use the following definition of "finite dimensional":

Definition A vector space $V$ is finite dimensional if $V$ has a finite spanning set. Otherwise, $V$ is infinite dimensional.

Note: the definition does not presume that we know what "dimension" means.

One has the following elementary theorem about matrices:

Theorem 1 If a matrix $A$ has more columns than rows, then there exists an non-zero solution to $A x = 0$.

Proof: Row reduction.

The theorem has the following consequence:

Theorem 2 Suppose $V$ is finite dimensional with a spanning set of cardinality $n$. Then any linearly independent set in $V$ has cardinality $\le n$.

Proof. Suppose $\{v_1, \dots, v_n\}$ is a spanning set and $\{w_1, \dots, w_m\}$ is linearly independent. Since $\{v_1, \dots, v_n\}$ is spanning, there exist $a_{i, j}$ in the underlying field $K$ such that $w_j = \sum a_{i, j} v_i$. Let $A$ denote the matrix $(a_{i, j})$. If $m >n$, there exists $$x = \begin{bmatrix}x_1 \\ \vdots \\ x_m \end{bmatrix} \ne 0$$ such that $A x = 0$. Check that this implies $\sum_j x_j w_j = 0$, contradicting the linear independence of $\{w_1, \dots, w_m\}$.

With a tiny bit more effort, one can prove the existence of bases and that any two bases have the same cardinality:

Definition A basis of $V$ is a set that is both spanning and linearly independent.

Theorem 3 A finite dimensional vector space has a basis.

Sketch: Choose a spanning set of minimal cardinality and show that it is also linearly independent.

Theorem 4 Any two bases of a finite dimensional vector space have the same cardinality.

Proof. Follows from Theorem 2.

Definition The cardinality of a basis is called the dimension of $V$.

Theorem Let $V$ be a finite dimensional vector space with dimension $n$. Then any spanning set of $V$ has cardinality $\ge n$ and any linearly independent subset of $V$ has cardinality $\le n$.

Proof. Follows from Theorem 2.

So: assume that for every $$n$$ there exists a linearly independent subset $$S_n$$ of $$V$$ of cardinality $$n$$. Show that there exists a countable subset of $$V$$ that is linearly independent.

Consider $$S'_n = S_1 \cup \ldots \cup S_n$$. Note that $$S'_1\subset S'_2 \subset \ldots$$ and inside every $$S'_n$$ there exists a linearly independent subset of cardinality $$n$$.

Now start an inductive construction. Inside $$S'_1$$ there exists a linearly independent subset $$S''_1$$ of cardinality $$1$$. Suppose that we have constructed $$S''_n$$ inside $$S'_n$$ of cardinality $$n$$. Then we can enlarge it to $$S''_{n+1}$$ inside $$S'_{n+1}$$ of cardinality $$n+1$$. We obtain $$S''_1 \subset S''_2 \subset S''_3 \subset \ldots$$ every $$S''_n$$ is linearly independent of cardinality $$n$$. We get $$S''=\cup_n S''_n$$ linearly independent and countable.