Suppose we have a homogeneous quadratic equation of three variables $w_1$, $w_2$, and $w_3 \in \mathbb{R}$ as follows:

$$W^TAW=0.$$ where $W=[w_1,w_2,w_3]^T$ and $A$ is a non-singular $3\times 3$ matrix. I was wondering if this equation can describe a shape rather than a translate-deformed-rotated sphere passing through origin of $\mathbb{R}^3$. Could it be a point, line, surface, or the union and intersection of these? If yes, how to characterize such shapes based on the properties of matrix $A$.


1 Answer 1


We will work in $\mathbb{R}^d$. We will denote $$w^T A w = \left\langle w, Aw \right\rangle,$$where $\left\langle \cdot , \cdot \right\rangle$ is the dot product and for $r \in \mathbb{R}$, $$S(A,r) = \left\{ x \in \mathbb{R}^d \: \middle| \: \left\langle w, Aw \right\rangle = r \right\}.$$ Now, what $S(A,r)$ is like, depends on the choice of $A$ and $r$. We will restrict to the case where the eigenvectors of $A$ form an orthonormal basis for $\mathbb{R}^d$ and derive some easily available things.

Note first that, for all $A$, $\left\{ 0 \right\} \subset S(A,0)$ (also note that if $x \in S(A,0)$, $\lambda x \in S(A,0)$ for all $\lambda \in \mathbb{R}$.) Let's first assume that $A$ has eigenvectors $e_1, \dots , e_d$ that form an orthonormal basis for $\mathbb{R}^d$ and that the eigenvalues $\lambda_1, \dots \lambda_d$ are positive. Then $$\left\langle w, Aw \right\rangle = \left\langle \sum \limits_{i=1}^d w_i e_i, \sum \limits_{i=1}^d w_i A e_i \right\rangle = \left\langle \sum \limits_{i=1}^d w_i e_i, \sum \limits_{i=1}^d w_i \lambda_i e_i \right\rangle = \sum \limits_{i=1}^d \lambda_i w_i^2. $$

Now $w \in S(A,r)$ iff $$\sum \limits_{i=1}^d \lambda_i w_i^2=r,$$ which, for $r>0$ is the surface of an ellipsoid. For $r=0$ you have that $S(A,0) = \left\{ 0 \right\}$. For $r < 0$ you have $S(A,r) = \varnothing$.

Now, if $A$ is as above, but has both positive and negative eigenvalues, things are more complicated. The equation $$\sum \limits_{i=1}^d \lambda_i w_i^2=r$$ may now describe shapes other than ellipsoids. Take e.g. $$A=\left( \begin{array}{cc} 1 & 0 \\ 0 &-1 \end{array} \right).$$Now $S(A,0)$ is a pair of intersecting lines and $S(A, 3)$ is a hyperbola.

If the eigenvalues of $A$ are allowed to be zero, $A$ has a null space $N_A$ with nonzero dimension. Clearly $N_A \subset S(A,0),$ so now you may have lines, planes etc. If $A = 0$, $S(A,0) = \mathbb{R}^d$.

  • $\begingroup$ Actually, matrix $A$ is the exponential of another matrix, i.e. $A=e^B$; so, in this case, all the eigenvalues of $A$ are positive, thus the quadratic form describes an ellipsoid. $\endgroup$
    – Saj_Eda
    May 23, 2016 at 20:50
  • $\begingroup$ Ok! But, as it turns out, in general you may have different things. $\endgroup$ May 23, 2016 at 20:52
  • $\begingroup$ Fine, thanks for your answer. $\endgroup$
    – Saj_Eda
    May 23, 2016 at 20:57
  • $\begingroup$ One more thing, what if an eigenvalue is non-real? $\endgroup$
    – Saj_Eda
    May 23, 2016 at 22:32
  • $\begingroup$ Well I'm not instantly sure. If the eigenvectors can be chosen orthonormal, the equation $\sum \limits_{i=1}^d \lambda_i w_i^2=r$ will still be valid. I'm not sure which types of sets they describe. $\endgroup$ May 24, 2016 at 6:27

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