- #1
SMA_01
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- 0
Let β be a basis for ℝ over Q (the set of all rational numbers) and let a[itex]\in[/itex]ℝ, a≠1.
Show that aβ={ay|y[itex]\in[/itex]β} is a basis for ℝ over Q for all a≠0.
So I need to show (1) Linear independence, and (2) spanning. I am a little confused, especially because the dimension for the vector space is uncountably infinite.
Here is what I have for linear independence:
I took a finite subset of aβ, {az|z[itex]\in[/itex]S[itex]\subset[/itex]β, |S|<|ℝ|} (not sure if I wrote that correctly), so if
[itex]\sum_{z\in S}[/itex]az=0
then z must be zero since a≠0 (initial assumption). Therefore, it follows that aβ is linearly independent.
Is this correct?
I don't know how to approach spanning.
Thanks in advance.
Show that aβ={ay|y[itex]\in[/itex]β} is a basis for ℝ over Q for all a≠0.
So I need to show (1) Linear independence, and (2) spanning. I am a little confused, especially because the dimension for the vector space is uncountably infinite.
Here is what I have for linear independence:
I took a finite subset of aβ, {az|z[itex]\in[/itex]S[itex]\subset[/itex]β, |S|<|ℝ|} (not sure if I wrote that correctly), so if
[itex]\sum_{z\in S}[/itex]az=0
then z must be zero since a≠0 (initial assumption). Therefore, it follows that aβ is linearly independent.
Is this correct?
I don't know how to approach spanning.
Thanks in advance.