0
$\begingroup$

If functions $f_1,\cdots f_n$ Are linearly independent in $C^{n-1}[a,b]$then they also are in $C[a,b]$.

But why is there a difference in (the definition of) linear independence for $C[a,b]$ vs $C^{n-1}[a,b]$?

The functions are the same, and the coefficients multiplying the functions for the linear combination are scalars, so where does the universal set (e.g. $C[a,b]$) come into play?

$\endgroup$
2
$\begingroup$

There is no difference. More generally, if $V$ is a subspace of a vector space $W$ and if $S\subset V$, then $S$ is linearly independent in $W$ if and only if it is linearly independent in $V$. This follows from the definition of linear independence.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.