Say we have an inner product such that $\{(1,0,0),(1,1,0),(1,1,1)\}$ is an orthonormal basis. I'm asked to find
- $||(1,2,3)||$
- An orthonormal basis for $U=$span$\{(0,1,0),(0,0,2)\}$
- The orthonormal projection of $v=(3,2,1)$ on $U$.
Now approaching this problem, I calculated 1 using the following method: We can write $(1,2,3)$ as $-1(1,0,0)-1(1,1,0)+3(1,1,1)$. We also know that for orthonormal basis, we can also write: $(1,2,3)=\langle(1,2,3),(1,0,0)\rangle(1,0,0)+\langle(1,2,3),(1,1,0)\rangle(1,1,0)+\langle(1,2,3),(1,1,1)\rangle(1,1,1)$
So we can infer the inner product of $\langle(1,2,3),(1,0,0)\rangle=1, \langle(1,2,3),(1,1,0)\rangle=1, \langle(1,2,3),(1,1,1)\rangle=3$
Using linearity of inner product we can then infer: $||(1,2,3)||^2=\langle(1,2,3),(1,2,3)\rangle=-1\langle(1,2,3),(1,0,0)\rangle-1\langle(1,2,3),(1,1,0)\rangle+3\langle(1,2,3),(1,1,1)\rangle=11$
Using the same approach I found that the orthonormal basis of $U$ is $B=\{\frac{1}{\sqrt2}(0,1,0),\frac{1}{\sqrt8}(0,1,2)\}$.
First of all - is my approach right? And secondly I'm having trouble with finding the projection, for some reason it always comes out $(0,0,0)$ which feels pretty weird to me.