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Given the bilinear form: $$(x,y)= x_1y_1+2x_1y_2+2x_2y_1+6x_2y_2$$ find an orthogonal basis for it.

I am confused on this. I was thinking that if I find the eigenvalues and corresponding eigenvectors, then I could find the orthogonal basis by dividing each eigenvector by its length since they are orthogonal themselves. However my eigenvalues were very odd. Thank you for any help.

Jean Marie
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Conor Hallen
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  • What do eigenvalues have to do with this? Take any basis and orthonormalize it using Gram-Schmidt... – DonAntonio Mar 23 '17 at 08:51
  • ok well don, how do i do that from a bilinear form, like how do i get the vectors from a bilinear form? – Conor Hallen Mar 23 '17 at 08:52
  • @ConorHallen, gram-schdmit. – IAmNoOne Mar 23 '17 at 08:52
  • @ConorHallen What? You have for example the vector $;u_1:=\binom10;$ , which is part of a basis (in fact, of infinite basis). Orthonormalize it wrt. to the given bilinear form, meaning: find its "length" wrt. to the bil. form$$(u_1,u_1):=1+2\cdot0+2\cdot0+6\cdot0=1$$ and etc. – DonAntonio Mar 23 '17 at 08:56
  • @DonAntonio I think the OP wanted to say "orthogonal basis of eigenvectors" – TZakrevskiy Mar 23 '17 at 08:56
  • @nameless, how do i use gram schdmit in this case to find it, can you tell me how to get the vectors from the bilinear form so i can use gram schmidt, i dont know how to get the vectors from it to do so. thank you for replying – Conor Hallen Mar 23 '17 at 08:56
  • @TZakrevskiy It is just the same: an orthogonal basis can be orthonormalized wrt to the given bil. form. – DonAntonio Mar 23 '17 at 08:57
  • @DonAntonio of course. But the resulting orthonormalized basis would not consist of eigenvectors. – TZakrevskiy Mar 23 '17 at 08:59
  • @TZakrevskiy If we begin with an orthogonal basis of eigenvectors, the orthonormalized oned will also be eigenvectors. In general, of course, this won't happen. But what do we care, though? The OP simply wants an orthonormal basis... – DonAntonio Mar 23 '17 at 09:00

1 Answers1

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With $\;u_1:=\binom10\;$ we have $(u_1,u_2)=1$. Now let $\;u_2:=\binom ab\;$ .

Determine $a,b$ such that

$(u_2,u_2)=1$ and $(u_1,u_2)=0$.

Then you are done.

Fred
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  • thank for you the response, by doing that i got a^2 +b^2=1, and a+b.0 =0, thus a=0 which means b=1? do they seem right for a=0 and b=1? – Conor Hallen Mar 23 '17 at 09:20