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$u_1 = (1, -1)'$ and $u_2 = (1, 1)'$ are two vector of $R^2$. Endow $R^2$ with an inner product such that $||u_1|| = 1$ and $||u_2|| = 1$.

Well, honestly, I don't completely understand what the problem asks. Endow $R^2$ with inner product? Then I tried the inner product of $u_1$ $u_2$. So, $<u_1,u_2> $ $= (1,-1)*(1,1)=0$. Then two vectors are orthogonal. But I don't know to how to proceed. Do i have show that $||u_1|| = 1 $ and $||u_2|| = 1 $? if so, what theorem or formula should I use?

Thanks in advance!

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  • $\begingroup$ Do you mean endow such that $||u_{1}|| = 1$ and $||u_{2}|| = 1$? $\endgroup$ – JessicaK Oct 24 '14 at 4:02
  • $\begingroup$ It wants you to find your own inner product where the conditions hold: not just apply the standard one. $\endgroup$ – Johanna Oct 24 '14 at 4:03
  • $\begingroup$ @Jessica K Yes! I don't understand what it means. $\endgroup$ – needhelp Oct 24 '14 at 4:04
  • $\begingroup$ @Johanna any hints? Thanks in advance! $\endgroup$ – needhelp Oct 24 '14 at 4:05
  • $\begingroup$ Off the top of my head: try to define $\langle u, v \rangle = \frac{x_1x_2 + y_1y_2}{2}$. Is this an inner product? Does it satisfy the requirements? $\endgroup$ – Johanna Oct 24 '14 at 4:09
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An inner product is a function of two vectors into $\Bbb R$ that satisfies certain properties. You are asked to find a function $f((a,b),(c,d))$ that satisfies these. You can't have $u_1=1$ because $u_1$ and $1$ are different kinds of things. You can have $f(u_1,u_1)=1$, which is what you want. The required linearity is a powerful constraint. Express any two vectors in the basis of $u_1,u_2$ and you know their inner product.

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  • $\begingroup$ So I have to find my own vectors $u_1$ and $u_2$ that satisfy $||u_1||=1 $ and $||u_2|| =1? Thanks for your hints tho! $\endgroup$ – needhelp Oct 24 '14 at 4:34
  • $\begingroup$ You are given the vectors in the problem statement. Now take any vector $(a,b)$, express it as a linear sum of $u_1,u_2$ and use linearity. $\endgroup$ – Ross Millikan Oct 24 '14 at 4:36
  • $\begingroup$ I guess I got it now! If I apply homogeneity in first slot, I will get $(au_1,u_1) = a(u_1,u_1)$ then find the value of a to satisfy $||u_1|| = 1$ and same method applies to $u_2$. Am i right? $\endgroup$ – needhelp Oct 24 '14 at 4:50
  • $\begingroup$ That is correct. $\endgroup$ – Ross Millikan Oct 24 '14 at 4:51
  • $\begingroup$ Thanks a lot! I was completely lost before. $\endgroup$ – needhelp Oct 24 '14 at 5:06

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