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There is a theorem that states for a finite dimensional non zero vector space $V$ if {$D^1,D^2$} are bases for subspace $M$ of $V$ such that {$D^1$} and {$D^2$} are also a non zero set of vectors in $M$. then {$D^1$} and {$D^2$} have the same number of vectors which makes sense. I couldn't find anything about subspaces with infinite dimensions but I assumed that this won't hold true because, for example, if we have {${D^1}$} with infinite number of independent vectors and {${D^2}$} with finite number of independent vectors then {${D^1}$} and {${D^2}$} will form the bases for $M$ but the number of vectors in {$D^1$} and {$D^2$} are not the same. First, am I correct to assume this and second is my reasoning correct? thank you!

Perfectoid
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    I don't understand your question, although I think I know what you want to ask. For example: "if ${D^1,D^2}$ is a bases for a subspace $M$" --- Are $D^1$ and $D^2$ elements of $M$ or subsets of $M$? Part of the problem is that ${D^1,D^2}$ is a single mathematical object, and thus "is a" seems to be confirming that you're talking about a single mathematical object (e.g. a set containing two elements), but then "bases" is plural, which now contradicts my initial reading. Look back over what you've written -- proof/edit some -- and perhaps be more explicit with what the symbols represent. – Dave L. Renfro Oct 23 '21 at 19:54
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    Another example: "then {$D^1$} and {$D^2$} have the same number of vectors" --- ${D^1}$ is a set that contains exactly ONE element, and ${D^2}$ is a set that contains exactly ONE element, so obviously they have the same number of elements. I don't have to know what $D^1$ and $D^2$ are to conclude this, although from what you've said (but maybe did not intend to say), $D^1$ and $D^2$ are vectors. – Dave L. Renfro Oct 23 '21 at 19:59
  • @DaveL.Renfro thank you for the feedback. I have edited the question. forgive me because I have trouble with English language. perhaps I did not understand the concept. – Perfectoid Oct 23 '21 at 20:15
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    Maybe this is relevant: Proof that two basis of a vector space have the same cardinality in the infinite-dimensional case – Dave L. Renfro Oct 23 '21 at 20:21
  • @DaveL.Renfro It does help very much indeed and I see where I got it confused. Thank you! – Perfectoid Oct 23 '21 at 20:26

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