Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm given a matrix $$A = \begin{pmatrix}-2&2\\2&1\end{pmatrix}$$ and we're asked to sketch the curve $\underline{x}^T A \underline{x} = 2$ where I assume $x = \begin{pmatrix}x\\y \end{pmatrix}$. Multiplying this out gives $-2 x^2+4 x y+y^2 = 2$.

Also, I diagonalised this matrix by creating a matrix, $P$, of normalised eigenvectors and computing $P^T\! AP = B$. This gives $$B = \begin{pmatrix}-3&0\\0&2\end{pmatrix}$$ and so now multiplying out $\underline{x}^T B \underline{x} = 2$ gives $-3x^2 + 2y^2 = 2$.

Plugging these equations into Wolfram Alpha gives different graphs, can someone please explain what I'm doing wrong? Thanks!

share|improve this question
Care to explain why you think they should be the same? –  nbubis Apr 13 '12 at 15:57
Oh I thought the reduction of a quadratic form to one which had a diagonal matrix in an orthonomal basis gave essentially the same quadratic form just in a 'nicer' way? –  user26069 Apr 13 '12 at 16:11

2 Answers 2

up vote 1 down vote accepted

Suppose $x$ is a point on $x^T A x = 2$. Then, $x^T P P^T A P P^T x = 2 $ $\Rightarrow x^T P B P^T x=2$. Thus the curve $x^T B x =2 $ is just the curve $x^T A x = 2$ rotated by $P^T$.

share|improve this answer

A quadratic form is "equivalent" to it's diagonalized form only in the sense that these quadratic forms share an equivalence class, but the resulting polynomials are of course different (as you showed), otherwise we wouldn't need the equivalence classes!

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.