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Vector spaces/ Subspaces/ Linear span properties

Kind of confused with the properties of vector spaces and linear spans. Need some help with the following questions
Screenshot 2021-09-27 at 7.54.46 PM.png
(edited 2 years ago)
Original post by rick flick
Kind of confused with the properties of vector spaces and linear spans. Need some help with the following questions
Screenshot 2021-09-27 at 7.54.46 PM.png


Any thoughts yourself regarding the two parts? Anything you can deduce?
Reply 2
Alright for (i) I think I got it.
(i) Span not equal means x is not a linear combi of v and w. We know x is redundant, so it must be a linear combi involving u, meaning the statement is true

(ii) For this I gather y is non-zero and not a linear combi of u v w x. I'm not sure if the new set can be expressed as a
linear span if y is a fixed vector.
Original post by rick flick
Alright for (i) I think I got it.
(i) Span not equal means x is not a linear combi of v and w. We know x is redundant, so it must be a linear combi involving u, meaning the statement is true


Not sure what you mean by "x is redundant" here, but yes, you have the essence of it. Since you're not feeling competent with this yet, you might find it useful padding this out and writing it in terms of the mathematics and justifying each of those parts; somewhat more long winded.
(Though if it's just a 1 mark question, they wouldn't be expecting you to do all that)



(ii) For this I gather y is non-zero and not a linear combi of u v w x. I'm not sure if the new set can be expressed as a
linear span if y is a fixed vector.


Think of the axioms of a vector (sub)space; in particular the zero vector.
Reply 4
Remember that for a finite dimensional vector space, any basis has the same number of vectors. In an n n -dimensional space, any n n linearly independent vectors form a basis.
Original post by ghostwalker
Not sure what you mean by "x is redundant" here, but yes, you have the essence of it. Since you're not feeling competent with this yet, you might find it useful padding this out and writing it in terms of the mathematics and justifying each of those parts; somewhat more long winded.
(Though if it's just a 1 mark question, they wouldn't be expecting you to do all that)


Original post by B_9710
Remember that for a finite dimensional vector space, any basis has the same number of vectors. In an n n -dimensional space, any n n linearly independent vectors form a basis.

I'm a little uncomfortable with these answers; it seems to me that (i) is very close to (a simple case of) the Steinitz exchange lemma; looking at it like that, I feel on the one hand the "x is redundant" argument is too vague, and on the other, proving that "any basis has the same number of vectors" usually relies on something similar to (i) so is arguably circular.

Personally, I'd do an explicit Steinitz-like proof (explicitly showing uSpan{v,w,x}u \in \text{Span}\{v, w, x\}).

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