# Inner product and unit vector

$u_1 = (1, -1)'$ 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?

• Do you mean endow such that $||u_{1}|| = 1$ and $||u_{2}|| = 1$? – JessicaK Oct 24 '14 at 4:02
• It wants you to find your own inner product where the conditions hold: not just apply the standard one. – Johanna Oct 24 '14 at 4:03
• @Jessica K Yes! I don't understand what it means. – needhelp Oct 24 '14 at 4:04
• @Johanna any hints? Thanks in advance! – needhelp Oct 24 '14 at 4:05
• 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? – Johanna Oct 24 '14 at 4:09

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.
• So I have to find my own vectors $u_1$ and $u_2$ that satisfy $||u_1||=1$ and $||u_2|| =1? Thanks for your hints tho! – needhelp Oct 24 '14 at 4:34 • You are given the vectors in the problem statement. Now take any vector$(a,b)$, express it as a linear sum of$u_1,u_2$and use linearity. – Ross Millikan Oct 24 '14 at 4:36 • I guess I got it now! If I apply homogeneity in first slot, I will get$(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? – needhelp Oct 24 '14 at 4:50