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I have the bilinear form defined by the matrix $\begin{pmatrix} 1 & 0 & 1 \\ 0 & 2 & 1 \\ 1 & 1 & 1 \end{pmatrix} $. Now I should find an orthogonal basis. I started by using Gram-Schmidt with the standard basis $e_{ij}$.

$v_1 = e_1=\begin{pmatrix} 1\\0\\0 \end{pmatrix}$

$v_2 = e_2-\frac{}{||v_1||}v_1=\begin{pmatrix} 0\\1\\0 \end{pmatrix}$

$v_3 = e_3-\frac{}{||v_1||}v_1 - \frac{}{||v_2||}v_2=\begin{pmatrix} -1\\-1\\1 \end{pmatrix}$

Now if I test these with my bilinear form, $,$ both yield 0 but $$ yields -1. Looked through the calculations multiple times but can't find the error.

I heard from other students that the order matters when you want to get an orthogonal basis. I thought the order only matters when you want to have a specific basis but Gram-Schmidts always results in an orthogonal basis?

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    You should find an orthonormal basis ... **for what** ? – 2017-02-03
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    It should be $\frac{\langle v_1, e_2 \rangle}{\langle v_1, v_1 \rangle}$, not $\frac{\langle v_1, e_2 \rangle}{|v_1|}$ – 2017-02-03
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    That's the exercise I got from the lecture – 2017-02-03
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    @Smoenybfan lisyarus is right, you need to compute $\|v_2\|$ with respect to the bilinear form, since Euclidean norm is meaningless here. – 2017-02-03
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    Don't I divide by the length? Or is the length always defined by the bilinear form I'm currently working with? And what do I do in the case of a division with 0? – 2017-02-03
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    @Smoenybfan Just see for yourself: assuming that $v_2 = e_2 - c v_1$, suppose that $\langle v_1, v_2 \rangle = 0$ for some scalar $c$. Solve for $c$ and you'll find that it *must* be equal to $\langle v_1, e_2 \rangle / \langle v_1, v_1 \rangle$. – 2017-02-03
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    Either you didn't write the whole exercise or else you didn't understand it: the bilinear form **has** to do something with all this, right? Why would it be relevant to use the usual inner product in this question? And even if it were, the standard basis is *already orthonormal* wrt the standard basis, so what you're doing wouldn't change anything – 2017-02-03
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    @DonAntonio As I check the computations, I think the user is actually using the form to compute the $\langle \cdot,\cdot\rangle$ quantities, but is probably just not using it to compute the $|\cdot|$ quantities. – 2017-02-03
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    @rschwieb Perhaps so. – 2017-02-03
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    @Smoenybfan To answer the last part of your question: yes, order affects the specific orthogonal basis that you end up with, but you always get an orthogonal set. There will never be a division by zero so long as the starting set is linearly independent and the bilinear form is non-degenerate. – 2017-02-03
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    @ErickWong ah yes thanks a lot, now I see it. Makes lot more sense than always using the scalarproduct. I just checked the matrix where I got the division by zero and saw that I can't use Gram-Schmidt because it's not non-degenerate. – 2017-02-03
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    @Smoenybfan You can still find an orthogonal basis for a degenerate form, it's just that some of the members necessarily have length zero. I think in this case it's better to find an orthogonal basis for the kernel of the form, then use regular gram-schmidt on the orthogonal complement (where the form is necessarily nondegenerate) – 2017-02-03
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    @Smoenybfan See [theorem 9 in Kaplansky's liner algebra book](https://books.google.com/books?id=nUcUYHrYJtgC&printsec=frontcover&dq=kaplansky+linear+algebra+second+course&hl=en&sa=X&ved=0ahUKEwjZu-aglPTRAhXn1IMKHfdPD0AQ6AEIJTAA#v=onepage&q=orthogonal&f=false) to see what I mean. – 2017-02-03

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