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I've read here that any quadratic form over $k^n$ with $n\in\mathbb{N}$ and $\text{char}(k)\neq 2$ is diagonalizable.

But considering $k=\mathbb{Q}$, which has characteristic $0$, the quadratic form $2x^2+2xy+y^2$ defined in $k^2$ is not diagonalizable, right?

What am I missing?

rmdmc89
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    $2x^2+2xy+y^2=x^2+(x+y)^2$ – Angina Seng Sep 13 '17 at 16:44
  • @LordSharktheUnknown, I've just realized my confusion. I was thinking about the eigenvalue decomposition of the matrix $\begin{bmatrix} 2 & 1 \ 1 & 1\ \end{bmatrix}$, which isn't possible in $\mathbb{Q}$ for this case. – rmdmc89 Sep 13 '17 at 17:18
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    Diagonalizing a quadratic form is easier than diagonalizing a matrix. The relevant matrix transformation is $M \mapsto QMQ^T$, not $M \mapsto QMQ^{-1}$, and in particular it does not preserve eigenvalues. – Qiaochu Yuan Sep 13 '17 at 18:07
  • @QiaochuYuan, what's the algorithm for $M\mapsto QMQ^{T}$? – rmdmc89 Sep 13 '17 at 19:50
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    I don't understand your question. Are you asking how to diagonalize quadratic forms? Basically the idea is to repeatedly complete the square, like Lord Shark did above. – Qiaochu Yuan Sep 13 '17 at 22:28
  • @QiaochuYuan, I was wondering about the general case for $k^n$. For example, take $n=3$ and the quadratic form $z^2-xy$. Completing the square doesn't apply here, right? – rmdmc89 Sep 13 '17 at 22:58
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    There you can instead "produce a square": $xy = \frac{(x + y)^2 - (x - y)^2}{4}$. – Qiaochu Yuan Sep 13 '17 at 23:32

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