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In the text i am referring for Linear Algebra , following definition for Infinite dimensional vector space is given .

The Vector Space V(F) is said to be infinite dimensional vector space or

infinitely generated if there exists an infinite subset S of V such that L(S) = V.

I am having following questions which the definition fails to answer :-

  1. Let us say i have an infinite dimensional vector space I(F) .If S is the infinite subset that spans I(F) , can i say that S is Linearly Independent . If so , why ?
  2. Does that mean any subset that spans every infinite dimensional vector space is Linearly Independent.
  3. On similar lines , can i say that Basis will exist for each Infinite Dimensional Vector Space and it is nothing but the subset that spans over vector space ?
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  • $\begingroup$ What's $L(S){}$? $\endgroup$ – Angina Seng Sep 24 '20 at 14:09
  • $\begingroup$ L(S) is Linear Combination of S over the field F. Here , S is a subset of V. Sorry for not mentioning. $\endgroup$ – llecxe Sep 24 '20 at 14:13
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    $\begingroup$ "If $S$ is the infinite subset that spans..." The article the carries some connotation in English that you think it is the only possible object that fits that description. Vector spaces in general can have many different spanning sets. You should have used the article a or an here instead implying you are aware that there could be others. Now... a spanning set does not need to be linearly independent. A basis does. Don't confuse these. As for your supposed definition you quoted, it is missing the condition that $S$ be linearly independent. $\endgroup$ – JMoravitz Sep 24 '20 at 14:15
  • $\begingroup$ Consider... $\Bbb R^2$ is equal to the span of the vectors $\{(0,1),(1,0),(2,0),(3,0),(4,0),(5,0),\dots\}$ yet $\Bbb R^2$ is 2-dimensional despite there being infinitely many vectors in the set used for the spanning set. It just so happened that this choice of spanning set was incredibly redundant and not formed using linearly independent vectors. $\endgroup$ – JMoravitz Sep 24 '20 at 14:16
  • $\begingroup$ Can we say any infinite dimensional vector space set will have a basis (whether finite or infinite ) for sure ? $\endgroup$ – llecxe Sep 24 '20 at 14:29
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Assuming that $L(S)$ denotes the linear hull (or span) of $S$, which is the set of all linear combinations that can be formed from finitely many elements in $S$, the statement you gave is wrong:

Note that $V=\mathbb R^n$ is a finite dimensional vector space over $\mathbb R$, yet for the infinite set $S=\mathbb R^n$, we have $L(S)=V$.

The correct definition is that $V$ is infinite dimensional if there exists no finite $S$ such that $L(S)=V$. Or equivalently (by the axiom of choice) if there exists an infinite set $S$ that is linearly independent and satisfies $L(S)=V$. That is, $V$ is infinite dimensional if it admits no finite basis, which is equivalent to admitting an infinite basis.

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  • $\begingroup$ How does V having no finite basis is equivalent to an infinite basis unless the linear independence of that infinite set is proved ? $\endgroup$ – llecxe Sep 24 '20 at 14:27
  • $\begingroup$ A basis is a linearly independent spanning set by definition. Every vector space admits a basis by the axiom of choice, so if there is no finite basis, there has to be an infinite one. $\endgroup$ – Christoph Sep 24 '20 at 14:29
  • $\begingroup$ Let W = {0} be subset of vector space V(F) ,where 0 is the identity element. Clearly W is a subspace of V(F). What is the basis of W? $\endgroup$ – llecxe Sep 24 '20 at 15:32
  • $\begingroup$ $W=\{0\}$ is a $0$-dimensional subspace and accordingly there's only one basis of $W$: the empty set $\varnothing$. $\endgroup$ – Christoph Sep 24 '20 at 16:25

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