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http://i.stack.imgr.com/QfTyD.png

Okay, I've done part (i) but I'm stumped on part (ii) and how I can show that.

Any help would be appreciated please.

user109331
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1 Answers1

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Suppose

$$\sum_{k=1}^n a_ke_k=0\implies\;\forall\,j=1,2,...,n\;,\;:$$

$$0=\left\langle \sum_{k=1}^na_ke_k\,,\,e_j\right\rangle=\sum_{k=1}^na_k\langle e_k\,,\,e_j\rangle=a_j$$

and we're done...

DonAntonio
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  • Thankyou for this, but can you tell me what the variable "a" represents? And also, how does this actually show that it is in fact linearly independent? So I know the workings behinds the answer to I understand it. – user109331 Mar 10 '14 at 05:02
  • $;a_i;$ are not variables but scalars, and you should read again carefully the definition of "linear independent set" in a vector space – DonAntonio Mar 10 '14 at 05:04
  • Just looked it up, makes sense now, thankyou! – user109331 Mar 10 '14 at 05:08
  • Any time, @user109331...this is stuff you can't allow yourself to forget if you're going to mess up with Hilbert, Banach and stuff spaces... – DonAntonio Mar 10 '14 at 05:10