Does c(t) = ((1+t,0),(0,1+t)) Provide a Counterexample on p.85 of Spivak's Book?

[zhentil]

Please forgive any stupid mistakes I've made.

On p.85, 4-5:

If $$c: [0,1] \rightarrow (R^n)^n$$ is continuous and each $$(c^1(t),c^2(t),...,c^n(t))$$ is a basis for $$R^n$$, prove that
$$|c^1(0),...,c^n(0)| = |c^1(1),...,c^n(1)|$$.

Maybe I'm missing something obvious, but doesn't $$c(t) = ((1+t,0),(0,1+t))$$ provide a counterexample to the statement when n=2?

 

[mathwonk]

have you read the section preceding the problem, specifically the definition of the notation [v1,...vn]? the bracket notation refers to the orientation defined by the basis. so the problem is to prove that a continuously varying basis cannot change orientation. this is easy since the determinant cannot change sign without being zero in between.

you have misrepresented the bracket notation as absolute value, at least according to my copy of spivak.

 

[zhentil]

I see. I have a (poor) photocopy of this section, and I can't distinguish between absolute value and the brackets you mentioned. Thank you for clarifying.

 

[mathwonk]

i suggest you buy one. then the author makes a few bucks, or maybe a nickel, i believe he told me once.

 

[Mystic998]

Spivak most definitely deserves some of your money. Possibly most of it.