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Consider the quadratic form given by the matrix below (in the canonical basis) \begin{pmatrix} 1 & 1 & -1\\ 1 & 1 & 3\\ -1 & 3 & 1 \end{pmatrix} Find an orthonormal basis of it and find its signature.


First I calculated the eigenvalues, which are $4, \frac{-1+ \sqrt{17}}{2}, \frac{-1-\sqrt{17}}{2}$. Then I calculated the eigenvectors associated to $4$ and $\frac{-1+ \sqrt{17}}{2}$ and normalized them, which gave me

\begin{align} e_1 &= \frac{1}{\sqrt{2}}\begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}, \quad e_2 = \frac{\sqrt{2}}{\sqrt{17+3\sqrt{17}}}\begin{bmatrix} -\frac{3+\sqrt{17}}{2} \\ -1 \\ 1 \end{bmatrix} \end{align}

And the third vector of the basis I want to be orthogonal to $e_1$ and $e_2$, so

$$e_3 = \frac{1}{\sqrt{17+3\sqrt{17}}} e_1 \wedge e_2 =\frac{1}{\sqrt{17+3\sqrt{17}}} \begin{bmatrix} 2 \\ -\frac{3+\sqrt{17}}{2} \\ \frac{3+\sqrt{17}}{2} \end{bmatrix} $$

I can't detail the calculations because they are very big. Perhaps someone can confirm the results. For the signature I know that the two possibilities are $(0,3)$ and $(2,1)$ but I don't know how to find the right one.

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  • $\begingroup$ Wolfram alpha tells me it's correct, although different normalizations are chosen: check it here $\endgroup$ Mar 24, 2022 at 11:30
  • $\begingroup$ As for the signature: you've already done it! Two positive eigenvalues and a negative one. $\endgroup$ Mar 24, 2022 at 11:31
  • $\begingroup$ @topolosaurus Wolfram did not normalize anything. $\endgroup$
    – KBS
    Mar 24, 2022 at 11:32
  • $\begingroup$ I don't understand your use of the cross product. That will give you a vector orthogonal to $e_1$ and $e_2$ for the euclidean structure, not the given quadratic form (unless you're lucky). $\endgroup$ Mar 24, 2022 at 11:59
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    $\begingroup$ @CaptainLama The cross product returns a vector that is orthogonal to $e_1$ and $e_2$. Note that the eigenvectors of a symmetric matrix are orthogonal. Note that the norm of the cross product is also one in this case. So, this means that it must be the unit-norm remaining eigenvector. $\endgroup$
    – KBS
    Mar 24, 2022 at 12:28

2 Answers 2

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This business with eigenvalues and eigenvectors is not how you diagonalize a quadratic form. It will give a correct result if done correctly, but is way too long and computationally painful. Simply use a "complete the square" method. In the canonical coordinates, your quadratic form is $$x^2 + y^2+z^2+2xy-2xz+6yz.$$ Now let us complete the squares: $$\begin{align} & x^2 + y^2+z^2+2xy-2xz+6yz \\ &= (x^2+2xy +y^2) + z^2-2xz+6yz \\ &= (x+y)^2 + z^2 -2xz+6yz \\ &= (x+y)^2 + z^2 -2z(x-3y) \\ &= (x+y)^2 + (z-(x-3y))^2 - (x-3y)^2. \end{align}$$ This means that the quadratic form is simply $(x')^2+(y')^2-(z')^2$ with the change of coordinates $$x'=x+y,\quad y'=-x+3y+z,\quad z'=x-3y.$$ This already clearly shows that the signature is $(2,1)$.

So to find your orthogonal basis, you just have to invert the matrix $$\begin{pmatrix} 1 & 1 & 0 \\ -1 & 3 & 1 \\ 1 & -3 & 0 \end{pmatrix}$$ and the basis will be given by the columns of the inverse. No need to compute eigenvalues or eigenvectors with complicated expressions. This gives $$\frac{1}{4}\begin{pmatrix} 3 & 0 & 1 \\ 1 & 0 & -1 \\ 0 & 4 & 4 \end{pmatrix},$$ so an orthonormal basis is $e_1 = \begin{bmatrix} 3/4 \\ 1/4 \\ 0\end{bmatrix}$, $e_2 = \begin{bmatrix} 0 \\ 0 \\ 1\end{bmatrix}$ and $e_3 = \begin{bmatrix} 1/4 \\ -1/4 \\ 1\end{bmatrix}$.

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Using SymPy:

>>> from sympy import *
>>> A = Matrix([[ 1, 1,-1],
                [ 1, 1, 3],
                [-1, 3, 1]])
>>> Q, D = A.diagonalize(normalize=True)
>>> D
Matrix([
[4,                 0,                 0],
[0, -1/2 + sqrt(17)/2,                 0],
[0,                 0, -sqrt(17)/2 - 1/2]])
>>> simplify(Q)
Matrix([
[        0, -sqrt(2)*(3 + sqrt(17))/(2*sqrt(3*sqrt(17) + 17)), sqrt(2)*(-3 + sqrt(17))/(2*sqrt(17 - 3*sqrt(17)))],
[sqrt(2)/2,                    -sqrt(2)/sqrt(3*sqrt(17) + 17),                    -sqrt(2)/sqrt(17 - 3*sqrt(17))],
[sqrt(2)/2,                     sqrt(2)/sqrt(3*sqrt(17) + 17),                     sqrt(2)/sqrt(17 - 3*sqrt(17))]])

Using function latex, we obtain

$$Q = \left[\begin{matrix}0 & - \frac{\sqrt{2} \left(3 + \sqrt{17}\right)}{2 \sqrt{3 \sqrt{17} + 17}} & \frac{\sqrt{2} \left(-3 + \sqrt{17}\right)}{2 \sqrt{17 - 3 \sqrt{17}}}\\\frac{\sqrt{2}}{2} & - \frac{\sqrt{2}}{\sqrt{3 \sqrt{17} + 17}} & - \frac{\sqrt{2}}{\sqrt{17 - 3 \sqrt{17}}}\\\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{\sqrt{3 \sqrt{17} + 17}} & \frac{\sqrt{2}}{\sqrt{17 - 3 \sqrt{17}}}\end{matrix}\right]$$

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