0
$\begingroup$

I'm given the inner product:

$\bigl\langle(x_1,x_2,x_3),(y_1,y_2,y_3)\bigr\rangle:=3x_1y_1+x_1y_3+y_1x_3+x_2y_2+2x_3y_3$

And I'm asked to find the orthonormal basis in respect of the above inner product that results from the normal basis $B=(e_1,e_2,e_3),$ $e_1=(1,0,0),e_2=(0,1,0), e_3=(0,0,1)$

My thought was to use the Gram-Schmidt process but use the given inner product to calculate the projections. However when I did so I ended up with the same vectors $e_1,e_2,e_3$ so the exact same basis...

I'm I doing somthing wrong or does the basis just stay the same?

$\endgroup$
  • 1
    $\begingroup$ No, e.g. $\langle e_1,e_1\rangle=3$ so $e_1$ will have to induced norm $\sqrt3$ instead of $1$. Also, we won't have $\langle e_1,e_3\rangle=0$, so they are not $\langle, \rangle$-orthogonal. $\endgroup$ – Berci May 6 '18 at 10:34
2
$\begingroup$

You are doing something wrong, and you are doing it from the start. Note that $\bigl\langle(1,0,0),(1,0,0)\bigr\rangle=3$ and that therefore the first vector that you should have got was $\frac1{\sqrt3}(1,0,0)$.

$\endgroup$
  • $\begingroup$ Ooooh yes! You are absolutely right I totally forgot to apply it to the first vector... Thank you $\endgroup$ – VakiPitsi May 6 '18 at 10:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.