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The task is to convert a quadric equation to a normal form.

$Q \equiv 3x_1^2 + 3x_2^2+3x_3^2+ 2x_1x_2+ 2x_1x_3 + 2x_2x_3+4x_1+4x_2+2x_3 +1=0$

What I've done is the following:

  1. Create a matrix A from the variables with degree 1 and 2

$$\begin{bmatrix} x_1 & x_2 & x_3\\ 3 & 1 & 1 & x_1\\ 1 & 3 & 1& x_2\\ 1 & 1 & 3& x_3\\ \end{bmatrix}$$

  1. Determine the determinant of $A- \lambda I_3 = -\lambda^3 +9\lambda^2 -24\lambda+20= (\lambda-2)^2(-\lambda + 5) $

  2. Determine the eigenvalues of $A- \lambda I_3$

$E_2= \operatorname{vect}\begin{pmatrix} 1\\ 1\\ 1\\ \end{pmatrix}$

$E_5= \operatorname{vect}\begin{pmatrix} 1\\ 1\\ 1\\ \end{pmatrix}$

but $\dim(E_2(A))+\dim(E_5(A)) \ne 3$. So a symmetrical matrix can't be constructed. Does this imply that there is no normal form? Can you maybe see if I made a mistake?

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  • $\begingroup$ Oops I forgot a term. I'll edit it. $\endgroup$ – Anonymous196 Jul 14 '18 at 22:38
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You are wrong about $E_2$ and $E_5$:$$E_5=\left\langle\begin{pmatrix}1\\1\\1\end{pmatrix}\right\rangle$$and$$E_2=\left\langle\begin{pmatrix}-1\\0\\1\end{pmatrix},\begin{pmatrix}-1\\1\\0\end{pmatrix}\right\rangle.$$So, $\dim E_5=1$ and $\dim E_2=2$.

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  • $\begingroup$ Oh, thanks for noticing that. Ok, now I can construct the symmetric matrix. $\endgroup$ – Anonymous196 Jul 14 '18 at 21:30
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You should’ve noticed for yourself that you’d done something wrong when you ended up with the same eigenvector for two different eigenvalues. $E_5$ is clearly correct by inspection—multiplying $A$ by this vector sums the rows—and let’s assume that you’re correct about that eigenspace being one-dimensional. By using the fact that the eigenspaces of a symmetric real matrix are mutually orthogonal, you can produce eigenvectors of $2$ with almost no work: $(-1,1,0)^T$, $(0,-1,1)^T$ and $(1,0,-1)^T$ are all clearly orthogonal to $(1,1,1)^T$. Any two of these, together with $E_5$, will do for diagonalizing the matrix. If you want an orthogonal basis, pick one of these and generate the other vector via a cross product, further normalizing everything if you’d like to end up with an orthonormal basis.

I’ll note that it looks like you’re trying to construct the canonical form rather than a normal form. The latter doesn’t really require finding eigenvalues/eigenvectors.

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  • $\begingroup$ Well yeah, I just dove in my textbook after two weeks of vacation without really remembering everything. I'll revise it soon enough. $\endgroup$ – Anonymous196 Jul 14 '18 at 22:37
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The normal form should be like this$$Q=x^TAx+x^Tb$$where$$A=\begin{bmatrix}3&2&2\\2&3&0\\2&0&3\end{bmatrix}$$and $$b=\begin{bmatrix}4\\4\\2\end{bmatrix}$$

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  • $\begingroup$ Huh, my A is different. But I was still busy constructing $x$. I think you should edit your $A$ $\endgroup$ – Anonymous196 Jul 14 '18 at 21:37

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