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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 $\operatorname{Span}(T)=S$?

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    $\begingroup$ If it was finite dimensional it would have a finite basis. $\endgroup$
    – copper.hat
    Commented Mar 5, 2018 at 22:42

3 Answers 3

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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.

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  • $\begingroup$ How can one assume that by assumption there is a set k+1<infinity elements such that the elements are linearly independent? Thanks $\endgroup$ Commented Mar 6, 2018 at 14:02
  • $\begingroup$ @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 $S_n$. $\endgroup$
    – Student
    Commented Mar 6, 2018 at 14:18
  • $\begingroup$ Sorry would you mind please explaining $\endgroup$ Commented Mar 6, 2018 at 14:41
  • $\begingroup$ @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. $\endgroup$
    – Student
    Commented Mar 6, 2018 at 14:43
  • $\begingroup$ 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) $\endgroup$ Commented Mar 18, 2018 at 15:42
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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.

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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.

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