# Finding an Orthonormal Basis using Gram Schmidt

Given the set of vectors $S=${${V_1=\binom{1}{4},V_2=\binom{4}{-4} }$} I am to find an orthonormal basis for $R^2$ using the Gram-Schmidt process. I've already worked it out and found the orthonormal basis $\beta$ to be $\beta=${$U_1= \binom{1\over{\sqrt{17}}}{4\over{\sqrt{17}}},U_2=\binom{5\over{\sqrt{17}}}{-5\over{\sqrt{17}}}$}. As far as I can tell I haven't made any arithmetic mistakes so I'm guessing it's the notation-picky software that's not accepting my answer. Could someone check my answer and see if matches please. If they don't match could you show me your work so I can see where I went wrong. Also if you have experience with the online linear algebra software through web-assign could you show me how I would enter this answer. Thank you in advance.

$$U_1 = V_1 = {1\choose4}$$

\begin{align} U_2 & =V_2-\text{proj}_{U_1}(V_2) \\ & ={4\choose{-4}}-\frac{U_1\cdot V_2}{||U_1||^2}U_1 \\ & ={4\choose{-4}}-\frac{-12}{17}{1\choose4} \\ & =\frac1{17}{80\choose{-20}} \\ \end{align}

$$\frac{U_1}{||U_1||} = \frac1{\sqrt{17}}{1\choose4}$$ $$\frac{U_2}{||U_2||} = \frac1{20\sqrt{17}}{80\choose{-20}}=\frac1{\sqrt{17}}{4\choose{-1}}$$

• This is incorrect. Look up the definition of projection. – TheBluegrassMathematician Apr 18 '14 at 1:30
• Sorry, I was wrong all over the place – user137794 Apr 18 '14 at 1:52
• Haha you're good. I have those days. Thank you for taking the time to try though. – TheBluegrassMathematician Apr 18 '14 at 1:55

Your $\beta$ appears not to be orthonormal. The inner product of $U_{1}$ and $U_{2}$ does not yield 1. There may be a problem in your construction. Please see that $< U_{1},U_{2}>= (\frac{1}{\sqrt{17}}*\frac{5}{\sqrt{17}}) + (\frac{4}{\sqrt{17}}*\frac{-5}{\sqrt{17}}) \neq 0$

• The inner product should rightly not equal because they are all orthogonal to each other therefore the inner product is qual to zero in which case I see that my answer is actually correct. – TheBluegrassMathematician Apr 18 '14 at 1:29